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;
const double pi = acos(0.0) * 2.0;
const long double eps = 1e-10;
const int step[8][2] = {{-1, 0}, {0, 1}, {1, 0},  {0, -1},
                        {-1, 1}, {1, 1}, {1, -1}, {-1, -1}};
template <class T>
inline T abs1(T a) {
  return a < 0 ? -a : a;
}
template <typename t, typename t1>
t min1(t a, t1 b) {
  return a < b ? a : b;
}
template <typename t, typename... arg>
t min1(t a, arg... arr) {
  return min1(a, min1(arr...));
}
template <typename t, typename t1>
t max1(t a, t1 b) {
  return a > b ? a : b;
}
template <typename t, typename... arg>
t max1(t a, arg... arr) {
  return max1(a, max1(arr...));
}
inline int jud(double a, double b) {
  if (abs(a) < eps && abs(b) < eps)
    return 0;
  else if (abs1(a - b) / max(abs1(a), abs1(b)) < eps)
    return 0;
  if (a < b) return -1;
  return 1;
}
template <typename t>
inline int jud(t a, t b) {
  if (a < b) return -1;
  if (a == b) return 0;
  return 1;
}
template <typename it, typename t1>
inline int RangeFind(t1 val, it a, int na, bool f_small = 1, bool f_lb = 1) {
  if (na == 0) return 0;
  int be = 0, en = na - 1;
  if (*a <= *(a + na - 1)) {
    if (f_lb == 0)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != 1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != -1)
          en = mid;
        else
          be = mid + 1;
      }
    if (f_small && jud(*(a + be), val) == 1) be--;
    if (!f_small && jud(*(a + be), val) == -1) be++;
  } else {
    if (f_lb)
      while (be < en) {
        int mid = (be + en + 1) / 2;
        if (jud(*(a + mid), val) != -1)
          be = mid;
        else
          en = mid - 1;
      }
    else
      while (be < en) {
        int mid = (be + en) / 2;
        if (jud(*(a + mid), val) != 1)
          en = mid;
        else
          be = mid + 1;
      }
    if (!f_small && jud(*(a + be), val) == -1) be--;
    if (f_small && jud(*(a + be), val) == 1) be++;
  }
  return be;
}
template <class T>
inline T lowb(T num) {
  return num & (-num);
}
inline int bitnum(unsigned int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(int nValue) { return __builtin_popcount(nValue); }
inline int bitnum(unsigned long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitnum(long long nValue) {
  return __builtin_popcount(nValue) + __builtin_popcount(nValue >> 32);
}
inline int bitmaxl(unsigned int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(int a) {
  if (a == 0) return 0;
  return 32 - __builtin_clz(a);
}
inline int bitmaxl(unsigned long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
inline int bitmaxl(long long a) {
  int temp = a >> 32;
  if (temp) return 32 - __builtin_clz(temp) + 32;
  return bitmaxl(int(a));
}
long long pow(long long n, long long m, long long mod = 0) {
  if (m < 0) return 0;
  long long ans = 1;
  long long k = n;
  while (m) {
    if (m & 1) {
      ans *= k;
      if (mod) ans %= mod;
    }
    k *= k;
    if (mod) k %= mod;
    m >>= 1;
  }
  return ans;
}
template <class t1, class t2>
inline void add(t1 &a, t2 b, int mod = -1) {
  if (mod == -1) mod = 1000000007;
  a += b;
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
}
template <class t>
void output1(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) cerr << arr[i] << ' ';
  cerr << endl;
}
template <class t>
void output2(t arr) {
  for (int i = 0; i < (int)arr.size(); i++) output1(arr[i]);
}
const int maxn = 600100;
struct SegmentNode {
  int be, en;
  unsigned long long add, sum, cnt;
};
struct SegmentTree {
  int l;
  SegmentNode tree[maxn * 4];
  inline int GetLeft(int no) { return no << 1; }
  inline int GetRight(int no) { return (no << 1) + 1; }
  inline int GetFa(int no) { return no >> 1; }
  inline SegmentTree() { l = 0; }
  void Build(int no, int l, int r) {
    if (l > r) r = l;
    if (l == r) {
      tree[no].be = tree[no].en = l;
      tree[no].sum = tree[no].cnt = 0;
      tree[no].add = -1;
      return;
    }
    tree[no].be = l;
    tree[no].en = r;
    int mid = (l + r) / 2;
    Build(GetLeft(no), l, mid);
    Build(GetRight(no), mid + 1, r);
    tree[no].add = -1;
    PushUp(no);
  }
  inline void PushUp(int no) {
    tree[no].sum = tree[GetLeft(no)].sum + tree[GetRight(no)].sum;
    tree[no].cnt = tree[GetLeft(no)].cnt + tree[GetRight(no)].cnt;
  }
  inline void PushDown(int no) {
    int le = GetLeft(no), ri = GetRight(no);
    tree[le].add = tree[no].add;
    tree[le].sum = tree[no].add * tree[le].cnt;
    tree[ri].add = tree[no].add;
    tree[ri].sum = tree[no].add * tree[ri].cnt;
    tree[no].add = -1;
  }
  void Upd(int l, int r, int no, unsigned long long ranadd, int cnt) {
    if (l <= tree[no].be && r >= tree[no].en) {
      tree[no].add = ranadd;
      if (cnt != -1) {
        tree[no].cnt = cnt;
      }
      tree[no].sum = ranadd * tree[no].cnt;
      return;
    }
    if (tree[no].add != -1 && tree[no].be != tree[no].en) PushDown(no);
    int mid = (tree[no].be + tree[no].en) >> 1;
    if (r >= tree[no].be && l <= mid) Upd(l, r, GetLeft(no), ranadd, cnt);
    if (r >= mid + 1 && l <= tree[no].en) Upd(l, r, GetRight(no), ranadd, cnt);
    PushUp(no);
  }
} sgt;
int kmp[maxn], nxt_have[maxn];
pair<int, int> st[maxn];
int lst;
char orig[maxn];
int val[maxn];
vector<int> fails[maxn];
int n;
void ExtendKMP(int loc) {
  if (loc == 0)
    kmp[loc] = -1;
  else
    kmp[loc] = kmp[loc - 1] + 1;
  while (kmp[loc] > 0 && orig[kmp[loc]] != orig[loc]) {
    fails[loc].push_back(kmp[loc]);
    kmp[loc] = kmp[kmp[loc] - 1] + 1;
  }
  if (kmp[loc] == -1 || orig[kmp[loc]] != orig[loc]) {
    kmp[loc] = -1;
    nxt_have[loc] = -1;
  } else {
    if (fails[kmp[loc]].size()) {
      nxt_have[loc] = kmp[loc];
    } else {
      nxt_have[loc] = nxt_have[kmp[loc]];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d", &n);
  sgt.Build(1, 0, n - 1);
  unsigned long long sum1 = 0, sum2 = 0, r26 = 0, r130 = 0;
  for (int i = 0; i < n; i++) {
    scanf("%s%d", orig + i, val + i);
    orig[i] -= 'a';
    orig[i] = (orig[i] + r26) % 26;
    val[i] = val[i] ^ (r130 & ((1ll << 30) - 1));
    ExtendKMP(i);
    st[lst] = make_pair(val[i], i);
    while (lst && st[lst - 1].first > val[i]) {
      lst--;
    }
    sgt.Upd(st[lst].second, i, 1, val[i], -1);
    if (orig[i] == orig[0]) {
      sgt.Upd(i, i, 1, val[i], 1);
    }
    for (int loc = i; loc > 0; loc = nxt_have[loc]) {
      for (auto it : fails[loc]) {
        int rloc = i - it;
        sgt.Upd(rloc, rloc, 1, 0, 0);
      }
    }
    st[lst] = make_pair(val[i], st[lst].second);
    lst++;
    sum1 += sgt.tree[1].sum;
    if (sum1 > 1000000000000ll) {
      sum2 += sum1 / 1000000000000ll;
      sum1 %= 1000000000000ll;
    }
    r26 += sgt.tree[1].sum;
    r26 %= 26;
    r130 += sgt.tree[1].sum;
    r130 &= (1ll << 30) - 1;
    if (sum2) {
      printf("%I64u%012I64u\n", sum2, sum1);
    } else {
      printf("%I64u\n", sum1);
    }
  }
  return 0;
}