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;
const long long MOD = 1e9 + 7;
long long tx[1 << 19], t[1 << 19];
struct SegtreeMOD {
  int n, h;
  SegtreeMOD() {}
  SegtreeMOD(int sz) {
    h = 32 - __builtin_clz(sz);
    n = 1 << h;
  }
  void assign(int x, long long v) {
    x += n;
    tx[x] = v % MOD;
    for (x /= 2; x; x /= 2) {
      tx[x] = (tx[x << 1] + tx[x << 1 | 1]);
      if (tx[x] >= MOD) {
        tx[x] -= MOD;
      }
    }
  }
  long long query(int l, int r) {
    long long ans = 0;
    for (l += n, r += n; l < r; l /= 2, r /= 2) {
      if (l % 2) {
        ans += tx[l++];
      }
      if (r % 2) {
        ans += tx[--r];
      }
    }
    return ans % MOD;
  }
};
struct Segtree {
  int n, h;
  Segtree() {}
  Segtree(int sz) {
    h = 32 - __builtin_clz(sz);
    n = 1 << h;
  }
  void assign(int x, long long v) {
    x += n;
    t[x] = v;
    for (x /= 2; x; x /= 2) {
      t[x] = t[x << 1] + t[x << 1 | 1];
    }
  }
  long long query(int l, int r) {
    long long ans = 0;
    for (l += n, r += n; l < r; l /= 2, r /= 2) {
      if (l % 2) {
        ans += t[l++];
      }
      if (r % 2) {
        ans += t[--r];
      }
    }
    return ans;
  }
  pair<long long, int> bs(int l, int r, long long goal, int x, int lv, int rv) {
    if (rv <= l || r <= lv) {
      return {0, -1};
    }
    if (l <= lv && rv <= r) {
      if (t[x] < goal) {
        return {t[x], -1};
      }
      if (rv - lv == 1) {
        return {t[x], lv};
      }
    }
    int mv = (lv + rv) / 2;
    auto res1 = bs(l, r, goal, x * 2, lv, mv);
    if (res1.second != -1) {
      return res1;
    }
    auto res2 = bs(l, r, goal - res1.first, x * 2 + 1, mv, rv);
    return {res2.first + res1.first, res2.second};
  }
  int bs(int l, int r, long long goal) {
    return bs(l, r, goal, 1, 0, n).second;
  }
};
Segtree weights;
SegtreeMOD wx;
int n, q;
long long a[200200];
long long query(int l, int r) {
  long long wtot = weights.query(l, r + 1);
  int med = weights.bs(l, r + 1, (wtot + 1) / 2);
  long long ans = ((weights.query(l, med) * 2 - wtot) % MOD) * a[med] -
                  wx.query(l, med) + wx.query(med, r + 1);
  ans = (ans % MOD + MOD) % MOD;
  return ans % MOD;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> q;
  weights = Segtree(n + 1);
  wx = SegtreeMOD(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    a[i] -= i;
  }
  for (int i = 1; i <= n; ++i) {
    long long w;
    cin >> w;
    weights.assign(i, w);
    wx.assign(i, w * a[i]);
  }
  while (q--) {
    int _a, b;
    cin >> _a >> b;
    if (_a > 0) {
      cout << query(_a, b) << "\n";
    } else {
      int id = -_a, nw = b;
      weights.assign(id, nw);
      wx.assign(id, nw * a[id]);
    }
  }
  return 0;
}