Problem: There is an infinite line consisting of cells. There are n boxes in some cells of this line. The i-th box stands in the cell a_i and has weight w_i. All a_i are distinct, moreover, a_{i - 1} < a_i holds for all valid i.

You would like to put together some boxes. Putting together boxes with indices in the segment [l, r] means that you will move some of them in such a way that their positions will form some segment [x, x + (r - l)].

In one step you can move any box to a neighboring cell if it isn't occupied by another box (i.e. you can choose i and change a_i by 1, all positions should remain distinct). You spend w_i units of energy moving the box i by one cell. You can move any box any number of times, in arbitrary order.

Sometimes weights of some boxes change, so you have queries of two types: 

  1. id nw — weight w_{id} of the box id becomes nw. 
  2. l r — you should compute the minimum total energy needed to put together boxes with indices in [l, r]. Since the answer can be rather big, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7. Note that the boxes are not moved during the query, you only should compute the answer. 



Note that you should minimize the answer, not its remainder modulo 10^9 + 7. So if you have two possible answers 2 ⋅ 10^9 + 13 and 2 ⋅ 10^9 + 14, you should choose the first one and print 10^9 + 6, even though the remainder of the second answer is 0.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the number of boxes and the number of queries.

The second line contains n integers a_1, a_2, ... a_n (1 ≤ a_i ≤ 10^9) — the positions of the boxes. All a_i are distinct, a_{i - 1} < a_i holds for all valid i.

The third line contains n integers w_1, w_2, ... w_n (1 ≤ w_i ≤ 10^9) — the initial weights of the boxes.

Next q lines describe queries, one query per line.

Each query is described in a single line, containing two integers x and y. If x < 0, then this query is of the first type, where id = -x, nw = y (1 ≤ id ≤ n, 1 ≤ nw ≤ 10^9). If x > 0, then the query is of the second type, where l = x and r = y (1 ≤ l_j ≤ r_j ≤ n). x can not be equal to 0.

Output

For each query of the second type print the answer on a separate line. Since answer can be large, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7.

Example

Input

5 8
1 2 6 7 10
1 1 1 1 2
1 1
1 5
1 3
3 5
-3 5
-1 10
1 4
2 5


Output

0
10
3
4
18
7

Note

Let's go through queries of the example: 

  1. 1\ 1 — there is only one box so we don't need to move anything. 
  2. 1\ 5 — we can move boxes to segment [4, 8]: 1 ⋅ |1 - 4| + 1 ⋅ |2 - 5| + 1 ⋅ |6 - 6| + 1 ⋅ |7 - 7| + 2 ⋅ |10 - 8| = 10. 
  3. 1\ 3 — we can move boxes to segment [1, 3]. 
  4. 3\ 5 — we can move boxes to segment [7, 9]. 
  5. -3\ 5 — w_3 is changed from 1 to 5. 
  6. -1\ 10 — w_1 is changed from 1 to 10. The weights are now equal to w = [10, 1, 5, 1, 2]. 
  7. 1\ 4 — we can move boxes to segment [1, 4]. 
  8. 2\ 5 — we can move boxes to segment [5, 8]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
inline void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const pair<T, U> &_p) {
  _out << _p.first << ' ' << _p.second;
  return _out;
}
template <typename T, typename U>
inline istream &operator>>(istream &_in, pair<T, U> &_p) {
  _in >> _p.first >> _p.second;
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const vector<T> &_v) {
  if (_v.empty()) {
    return _out;
  }
  _out << _v.front();
  for (auto _it = ++_v.begin(); _it != _v.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline istream &operator>>(istream &_in, vector<T> &_v) {
  for (auto &_i : _v) {
    _in >> _i;
  }
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const set<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const multiset<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const unordered_set<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const unordered_multiset<T> &_s) {
  if (_s.empty()) {
    return _out;
  }
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) {
    _out << ' ' << *_it;
  }
  return _out;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const map<T, U> &_m) {
  if (_m.empty()) {
    return _out;
  }
  _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')';
  for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) {
    _out << ", (" << _it->first << ": " << _it->second << ')';
  }
  return _out;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const unordered_map<T, U> &_m) {
  if (_m.empty()) {
    return _out;
  }
  _out << '(' << _m.begin()->first << ": " << _m.begin()->second << ')';
  for (auto _it = ++_m.begin(); _it != _m.end(); ++_it) {
    _out << ", (" << _it->first << ": " << _it->second << ')';
  }
  return _out;
}
const string FILENAME = "input";
const int MAXN = 200228;
const int M = 1000000007;
const long long base = 1000000000000000000LL;
struct fenwick {
  vector<long long> f;
  int nn;
  void init(int n) {
    f.resize(n, 0);
    nn = n;
  }
  void inc(int i, long long val) {
    for (int j = i; j < nn; j = (j | (j + 1))) {
      f[j] += val;
    }
  }
  long long getsum(int l) {
    if (l < 0) {
      return 0;
    }
    long long t = 0;
    for (int j = l; j >= 0; j = (j & (j + 1)) - 1) {
      t += f[j];
    }
    return t;
  }
  long long get(int l, int r) { return getsum(r) - getsum(l - 1); }
};
struct fenwick1 {
  vector<long long> f;
  int nn;
  void init(int n) {
    f.resize(n, 0);
    nn = n;
  }
  void inc(int i, long long val) {
    for (int j = i; j < nn; j = (j | (j + 1))) {
      f[j] += val;
      f[j] %= M;
    }
  }
  long long getsum(int l) {
    if (l < 0) {
      return 0;
    }
    long long t = 0;
    for (int j = l; j >= 0; j = (j & (j + 1)) - 1) {
      t += f[j];
      t %= M;
    }
    return t;
  }
  long long get(int l, int r) { return (getsum(r) - getsum(l - 1)) % M; }
};
int n, q;
int a[MAXN];
int w[MAXN];
fenwick wss;
fenwick1 wss1;
fenwick1 pref;
fenwick1 suff;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  wss.init(n);
  wss1.init(n);
  pref.init(n);
  suff.init(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    cin >> w[i];
    wss.inc(i, w[i]);
    wss1.inc(i, 1LL * w[i] * a[i] % M);
  }
  for (int i = 0; i < n; i++) {
    pref.inc(i, 1LL * w[i] * (i + 1) % M);
  }
  for (int i = 0; i < n; i++) {
    suff.inc(i, 1LL * w[i] * (n - i) % M);
  }
  for (int it = 0; it < q; it++) {
    int first;
    cin >> first;
    if (first < 0) {
      int second;
      cin >> second;
      first *= -1;
      first--;
      wss.inc(first, second - w[first]);
      wss1.inc(first,
               (1LL * second * a[first]) % M - (1LL * w[first] * a[first]) % M);
      pref.inc(first, (1LL * second * (first + 1)) % M -
                          1LL * w[first] * (first + 1) % M);
      suff.inc(first, (1LL * second * (n - first)) % M -
                          1LL * w[first] * (n - first) % M);
      w[first] = second;
    } else {
      int second;
      cin >> second;
      first--, second--;
      int l = first;
      int r = second;
      long long f = wss.get(first, second);
      while (l != r) {
        int mid = (l + r) >> 1;
        long long k = wss.get(first, mid);
        if (k >= f - k) {
          r = mid;
        } else {
          l = mid + 1;
        }
      }
      long long ans = 0;
      if (first < l) {
        long long t = wss1.get(first, l - 1);
        long long f = 1LL * a[l] * (wss.get(first, l - 1) % M) % M;
        f -= suff.get(first, l - 1);
        f += (wss.get(first, l - 1) * (n - l)) % M;
        f -= t;
        f %= M;
        if (f < 0) {
          f += M;
        }
        ans += f;
        ans %= M;
      }
      if (l + 1 <= second) {
        long long t = wss1.get(l + 1, second);
        long long f = 1LL * a[l] * (wss.get(l + 1, second) % M) % M;
        f += pref.get(l + 1, second);
        f -= (wss.get(l + 1, second) * (l + 1)) % M;
        t -= f;
        t %= M;
        if (t < 0) {
          t += M;
        }
        ans += t;
        ans %= M;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}