Problem: Fedya has a string S, initially empty, and an array W, also initially empty.

There are n queries to process, one at a time. Query i consists of a lowercase English letter c_i and a nonnegative integer w_i. First, c_i must be appended to S, and w_i must be appended to W. The answer to the query is the sum of suspiciousnesses for all subsegments of W [L, \ R], (1 ≤ L ≤ R ≤ i).

We define the suspiciousness of a subsegment as follows: if the substring of S corresponding to this subsegment (that is, a string of consecutive characters from L-th to R-th, inclusive) matches the prefix of S of the same length (that is, a substring corresponding to the subsegment [1, \ R - L + 1]), then its suspiciousness is equal to the minimum in the array W on the [L, \ R] subsegment. Otherwise, in case the substring does not match the corresponding prefix, the suspiciousness is 0.

Help Fedya answer all the queries before the orderlies come for him!

Input

The first line contains an integer n (1 ≤ n ≤ 600 000) — the number of queries.

The i-th of the following n lines contains the query i: a lowercase letter of the Latin alphabet c_i and an integer w_i (0 ≤ w_i ≤ 2^{30} - 1).

All queries are given in an encrypted form. Let ans be the answer to the previous query (for the first query we set this value equal to 0). Then, in order to get the real query, you need to do the following: perform a cyclic shift of c_i in the alphabet forward by ans, and set w_i equal to w_i ⊕ (ans \ \& \ MASK), where ⊕ is the bitwise exclusive "or", \& is the bitwise "and", and MASK = 2^{30} - 1.

Output

Print n lines, i-th line should contain a single integer — the answer to the i-th query.

Examples

Input


7
a 1
a 0
y 3
y 5
v 4
u 6
r 8


Output


1
2
4
5
7
9
12


Input


4
a 2
y 2
z 0
y 2


Output


2
2
2
2


Input


5
a 7
u 5
t 3
s 10
s 11


Output


7
9
11
12
13

Note

For convenience, we will call "suspicious" those subsegments for which the corresponding lines are prefixes of S, that is, those whose suspiciousness may not be zero.

As a result of decryption in the first example, after all requests, the string S is equal to "abacaba", and all w_i = 1, that is, the suspiciousness of all suspicious sub-segments is simply equal to 1. Let's see how the answer is obtained after each request:

1. S = "a", the array W has a single subsegment — [1, \ 1], and the corresponding substring is "a", that is, the entire string S, thus it is a prefix of S, and the suspiciousness of the subsegment is 1.

2. S = "ab", suspicious subsegments: [1, \ 1] and [1, \ 2], total 2.

3. S = "aba", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3] and [3, \ 3], total 4.

4. S = "abac", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] and [3, \ 3], total 5.

5. S = "abaca", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] , [1, \ 5], [3, \ 3] and [5, \ 5], total 7.

6. S = "abacab", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] , [1, \ 5], [1, \ 6], [3, \ 3], [5, \ 5] and [5, \ 6], total 9.

7. S = "abacaba", suspicious subsegments: [1, \ 1], [1, \ 2], [1, \ 3], [1, \ 4] , [1, \ 5], [1, \ 6], [1, \ 7], [3, \ 3], [5, \ 5], [5, \ 6], [5, \ 7] and [7, \ 7], total 12.

In the second example, after all requests S = "aaba", W = [2, 0, 2, 0].

1. S = "a", suspicious subsegments: [1, \ 1] (suspiciousness 2), totaling 2.

2. S = "aa", suspicious subsegments: [1, \ 1] (2), [1, \ 2] (0), [2, \ 2] ( 0), totaling 2.

3. S = "aab", suspicious subsegments: [1, \ 1] (2), [1, \ 2] (0), [1, \ 3] ( 0), [2, \ 2] (0), totaling 2.

4. S = "aaba", suspicious subsegments: [1, \ 1] (2), [1, \ 2] (0), [1, \ 3] ( 0), [1, \ 4] (0), [2, \ 2] (0), [4, \ 4] (0), totaling 2.

In the third example, from the condition after all requests S = "abcde", W = [7, 2, 10, 1, 7].

1. S = "a", suspicious subsegments: [1, \ 1] (7), totaling 7.

2. S = "ab", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), totaling 9.

3. S = "abc", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), [1, \ 3] ( 2), totaling 11.

4. S = "abcd", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), [1, \ 3] ( 2), [1, \ 4] (1), totaling 12.

5. S = "abcde", suspicious subsegments: [1, \ 1] (7), [1, \ 2] (2), [1, \ 3] ( 2), [1, \ 4] (1), [1, \ 5] (1), totaling 13.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = __int128;
using ld = long double;
template <typename T>
vector<T>& operator--(vector<T>& v) {
  for (auto& i : v) --i;
  return v;
}
template <typename T>
vector<T>& operator++(vector<T>& v) {
  for (auto& i : v) ++i;
  return v;
}
template <typename T>
istream& operator>>(istream& is, vector<T>& v) {
  for (auto& i : v) is >> i;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> v) {
  for (auto& i : v) os << i << ' ';
  return os;
}
template <typename T, typename U>
pair<T, U>& operator--(pair<T, U>& p) {
  --p.first;
  --p.second;
  return p;
}
template <typename T, typename U>
pair<T, U>& operator++(pair<T, U>& p) {
  ++p.first;
  ++p.second;
  return p;
}
template <typename T, typename U>
istream& operator>>(istream& is, pair<T, U>& p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
ostream& operator<<(ostream& os, pair<T, U> p) {
  os << p.first << ' ' << p.second;
  return os;
}
template <typename T, typename U>
pair<T, U> operator-(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first - b.first, a.second - b.second);
}
template <typename T, typename U>
pair<T, U> operator+(pair<T, U> a, pair<T, U> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
template <typename T, typename U>
void umin(T& a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
void umax(T& a, U b) {
  if (a < b) a = b;
}
mt19937 rnd = mt19937(934287);
template <typename T, vector<T>& v>
struct static_pointer {
  int ind = -1;
  static_pointer() {}
  static_pointer(int ind) : ind(ind) {}
  static_pointer(T* ptr) {
    v.push_back(*ptr);
    delete ptr;
    ind = v.size() - 1;
  }
  static_pointer(nullptr_t p) : ind(-1) {}
  template <typename... Args>
  static static_pointer<T, v> new_node(Args... args) {
    v.emplace_back(args...);
    return v.size() - 1;
  }
  T* operator->() const { return &v[ind]; }
  T& operator*() { return v[ind]; }
  operator bool() const { return ind != -1; }
  operator int() const { return ind; }
  bool operator==(nullptr_t p) const { return ind == -1; }
  bool operator==(int k) const { return ind == k; }
  bool operator!=(nullptr_t p) const { return ind != -1; }
  bool operator!=(int k) const { return ind != k; }
};
template <typename T, vector<T>& v>
ostream& operator<<(ostream& o, const static_pointer<T, v>& ptr) {
  return o << (int)ptr;
}
struct treapArNode;
vector<treapArNode> node_pool;
using treap_ar_node_ptr = static_pointer<treapArNode, node_pool>;
struct treapArNode {
  treap_ar_node_ptr left = nullptr;
  treap_ar_node_ptr right = nullptr;
  int value;
  int size;
  ll sm;
  bool is_set = false;
  int set;
  int ind;
  treapArNode(int ind, int _value) : value(_value), size(1), ind(ind) {
    sm = value;
  }
  void reset() {
    size = 1;
    sm = value;
    if (left != nullptr) {
      size += left->size;
      sm += left->sm;
    }
    if (right != nullptr) {
      size += right->size;
      sm += right->sm;
    }
  }
};
struct treapAr {
  treap_ar_node_ptr root = nullptr;
  treapAr() {}
  treapAr(treap_ar_node_ptr _root) { root = _root; }
  pair<treap_ar_node_ptr, treap_ar_node_ptr> split(treap_ar_node_ptr node,
                                                   int k) {
    push(node);
    if (node == nullptr) {
      return {nullptr, nullptr};
    }
    if (node->ind == k) {
      auto left = node->left;
      node->left = nullptr;
      node->reset();
      return {left, node};
    }
    if (node->ind > k) {
      auto ans = split(node->left, k);
      node->left = ans.second;
      node->reset();
      return {ans.first, node};
    } else {
      auto ans = split(node->right, k);
      node->right = ans.first;
      node->reset();
      return {node, ans.second};
    }
  }
  ll ask() {
    if (!root) return 0;
    return root->sm;
  }
  treap_ar_node_ptr merge(treap_ar_node_ptr a, treap_ar_node_ptr b) {
    push(a);
    push(b);
    if (a == nullptr) {
      return b;
    } else if (b == nullptr) {
      return a;
    }
    if (rnd() % (a->size + b->size) < a->size) {
      a->right = merge(a->right, b);
      a->reset();
      return a;
    } else {
      b->left = merge(a, b->left);
      b->reset();
      return b;
    }
  }
  void modify(treap_ar_node_ptr node, int val) {
    if (!node) return;
    node->is_set = true;
    node->set = val;
    node->value = val;
    node->sm = (ll)val * node->size;
  }
  void push(treap_ar_node_ptr node) {
    if (!node) return;
    if (node->is_set) {
      modify(node->left, node->set);
      modify(node->right, node->set);
      node->is_set = false;
    }
  }
  void insert(int i, int val) {
    treap_ar_node_ptr newNode = treap_ar_node_ptr::new_node(i, val);
    root = merge(root, newNode);
  }
  void set(int i, int val) {
    void(0);
    auto [L, R] = split(root, i);
    if (R) {
      push(R);
      modify(R, val);
    }
    root = merge(L, R);
  }
  treap_ar_node_ptr get(int i) {
    treap_ar_node_ptr node = root;
    while (node != nullptr) {
      if (node->left == nullptr) {
        if (i == 0) {
          return node;
        } else {
          --i;
          node = node->right;
        }
      } else {
        if (i == node->left->size) {
          return node;
        } else if (i < node->left->size) {
          node = node->left;
        } else {
          i -= node->left->size + 1;
          node = node->right;
        }
      }
    }
  }
  int operator[](int i) { return get(i)->value; }
  vector<int> toVector(treap_ar_node_ptr node) {
    if (node == nullptr) {
      return vector<int>({});
    }
    vector<int> vLeft = toVector(node->left);
    vector<int> vRight = toVector(node->right);
    vLeft.push_back(node->value);
    vLeft.insert(vLeft.end(), vRight.begin(), vRight.end());
    return vLeft;
  }
  vector<int> toVector() { return toVector(root); }
  int extract(int k) {
    auto p1 = split(root, k);
    auto p2 = split(p1.second, k + 1);
    int x = p2.first->value;
    root = merge(p1.first, p2.second);
    return x;
  }
  treap_ar_node_ptr cut(int l, int r) {
    auto p1 = split(root, l);
    auto p2 = split(p1.second, r + 1 - l);
    root = merge(p1.first, p2.second);
    return p2.first;
  }
};
string to_string(__int128 x) {
  if (x == 0) return "0";
  __uint128_t k = x;
  if (k == (((__uint128_t)1) << 127))
    return "-170141183460469231731687303715884105728";
  string result;
  if (x < 0) {
    result += "-";
    x *= -1;
  }
  string t;
  while (x) {
    t.push_back('0' + x % 10);
    x /= 10;
  }
  reverse(t.begin(), t.end());
  return result + t;
}
ostream& operator<<(ostream& o, __int128 x) { return o << to_string(x); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> next_good(n, 0);
  vector<int> v(n);
  string s;
  node_pool.reserve(1.5e6);
  vector<vector<int>> next_if(26, vector<int>(n, 0));
  ll prev_ans = 0;
  ll total = 0;
  int min_all = 2e9;
  ll other = 0;
  treapAr cur;
  vector<int> st;
  for (int i = 0; i < n; ++i) {
    char c;
    int val;
    cin >> c >> val;
    c = (c - 'a' + prev_ans) % 26 + 'a';
    val = val ^ (prev_ans & ((1ll << 30) - 1));
    min_all = min(min_all, val);
    v[i] = val;
    s.push_back(c);
    if (i == 0) {
      prev_ans = total = val;
      cout << val << '\n';
      continue;
    }
    {
      int x = i;
      while (x) {
        x = next_good[x - 1];
        if (next_if[s[x] - 'a'][i] == 0) next_if[s[x] - 'a'][i] = x + 1;
        if (s[i] == s[x]) {
          if (next_good[i] == 0) next_good[i] = x + 1;
          break;
        }
      }
      if (s.back() == s[0] && next_good[i] == 0) next_good[i] = 1;
    }
    for (int j = 0; j < 26; ++j) {
      if (next_if[j][i] != 0) continue;
      if (next_good[i] == 0) continue;
      next_if[j][i] = next_if[j][next_good[i] - 1];
    }
    s.back() = c;
    next_good[i] = next_if[s.back() - 'a'][i];
    for (int j = 0; j < 26; ++j) {
      if (s.back() == 'a' + j) continue;
      int x = i;
      x = next_if[j][x];
      while (x) {
        if (x <= 1) break;
        int v = cur.extract(i - x + 1);
        total += v;
        other -= v;
        x = next_good[x - 1];
      }
    }
    total += other;
    while (!st.empty() && v[st.back()] >= v[i]) st.pop_back();
    cur.set(st.empty() ? 0 : st.back() + 1, v[i]);
    other = cur.ask();
    if (s[i] == s[0]) {
      cur.insert(i, val);
      other += val;
    }
    prev_ans = total + other + min_all;
    total += min_all;
    cout << prev_ans << '\n';
    st.push_back(i);
  }
  return 0;
}