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;
using ll = long long;
constexpr ll MOD = 1000000007;
struct fenwick {
  int n;
  vector<ll> tree;
  bool modOut;
  fenwick(int n) : n(n), tree(n + 1) {}
  fenwick(vector<ll> init, bool modOut)
      : n(init.size()), tree(init.size() + 1), modOut(modOut) {
    for (int i = 0; i < n; ++i) add(i, init[i]);
  }
  int LSB(int i) { return (i & (-i)); }
  void add(int i, ll v) {
    for (int ci = i + 1; ci <= n; ci += LSB(ci)) {
      tree[ci] = (tree[ci] + v);
      if (modOut) tree[ci] %= MOD;
    }
  }
  ll getPrefix(int i) {
    ll ans = 0;
    for (int ci = i + 1; ci > 0; ci -= LSB(ci)) {
      ans += tree[ci];
      if (modOut) ans %= MOD;
    }
    return ans;
  }
  ll getRange(int i, int j) {
    ll ans = getPrefix(j) - getPrefix(i - 1);
    if (modOut) ans = (ans + MOD) % MOD;
    return ans;
  }
};
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  vector<ll> boxPos(n);
  for (int i = 0; i < n; ++i) scanf("%lld", &boxPos[i]);
  for (int i = 0; i < n; ++i) boxPos[i] -= i;
  vector<ll> weights(n);
  for (int i = 0; i < n; ++i) scanf("%lld", &weights[i]);
  vector<ll> weightedSum(n);
  for (int i = 0; i < n; ++i) {
    weightedSum[i] = (boxPos[i] * weights[i]) % MOD;
  }
  fenwick sWeights(weights, false), wbProd(weightedSum, true);
  for (int qta = 0; qta < q; ++qta) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x < 0) {
      x = -(x + 1);
      sWeights.add(x, -weights[x]);
      wbProd.add(x, (MOD - boxPos[x] * weights[x] % MOD) % MOD);
      weights[x] = y;
      sWeights.add(x, weights[x]);
      wbProd.add(x, boxPos[x] * weights[x] % MOD);
    } else {
      --x;
      --y;
      int lo = x, hi = y;
      ll rangewSum = sWeights.getRange(x, y);
      while (lo < hi) {
        int mid = (lo + hi) / 2;
        if (sWeights.getRange(x, mid) * 2 >= rangewSum)
          hi = mid;
        else
          lo = mid + 1;
      }
      ll centerPoint = boxPos[lo];
      ll ans = (sWeights.getRange(x, lo) % MOD * centerPoint % MOD -
                wbProd.getRange(x, lo) + MOD) %
               MOD;
      if (lo < y) {
        ans = (ans +
               (wbProd.getRange(lo + 1, y) -
                sWeights.getRange(lo + 1, y) % MOD * centerPoint % MOD + MOD) %
                   MOD) %
              MOD;
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}