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 int MAXN = 200000;
const int INF = (long long)1e9 + 5;
const long long MOD = (long long)1e9 + 7;
int N;
long long pos[MAXN + 5], w[MAXN + 5];
struct Segment_tree {
  long long sum0[4 * MAXN + 5], sum1[4 * MAXN + 5], psum[4 * MAXN + 5];
  void update(int p, int l, int r, int idx) {
    if (l == p && r == p) {
      sum0[idx] = w[p];
      sum1[idx] = w[p] * p % MOD;
      psum[idx] = pos[p] * w[p] % MOD;
      return;
    }
    int m = (l + r) >> 1;
    if (p <= m)
      update(p, l, m, idx << 1);
    else
      update(p, m + 1, r, idx << 1 | 1);
    sum0[idx] = (sum0[idx << 1] + sum0[idx << 1 | 1]);
    sum1[idx] = (sum1[idx << 1] + sum1[idx << 1 | 1]) % MOD;
    psum[idx] = (psum[idx << 1] + psum[idx << 1 | 1]) % MOD;
  }
  long long RSQ(long long *tar, int ql, int qr, int l, int r, int idx,
                bool mod) {
    if (qr < l || r < ql) return 0;
    if (ql <= l && r <= qr) return tar[idx];
    int m = (l + r) >> 1;
    long long ans = RSQ(tar, ql, qr, l, m, idx << 1, mod) +
                    RSQ(tar, ql, qr, m + 1, r, idx << 1 | 1, mod);
    if (mod) ans %= MOD;
    return ans;
  }
  int search(long long x, int l, int r, int idx) {
    if (l == r) return l;
    int m = (l + r) >> 1;
    if (sum0[idx << 1] > x) return search(x, l, m, idx << 1);
    return search(x - sum0[idx << 1], m + 1, r, idx << 1 | 1);
  }
} tree;
long long query(int l, int r) {
  long long sum = tree.RSQ(tree.sum0, l, r, 0, N, 1, false);
  long long pref = tree.RSQ(tree.sum0, 0, l - 1, 0, N, 1, false);
  int tar = tree.search(pref + sum / 2, 0, N, 1);
  assert(l <= tar && tar <= r);
  long long lpsum = tree.RSQ(tree.psum, l, tar, 0, N, 1, 1),
            rpsum = tree.RSQ(tree.psum, tar, r, 0, N, 1, 1);
  long long lsum0 = tree.RSQ(tree.sum0, l, tar, 0, N, 1, 1),
            rsum0 = tree.RSQ(tree.sum0, tar, r, 0, N, 1, 1);
  long long lsum1 = tree.RSQ(tree.sum1, l, tar, 0, N, 1, 1),
            rsum1 = tree.RSQ(tree.sum1, tar, r, 0, N, 1, 1);
  long long lans =
      (pos[tar] * lsum0 % MOD - lpsum - tar * lsum0 % MOD + lsum1) % MOD;
  long long rans =
      (rpsum - pos[tar] * rsum0 % MOD - rsum1 + tar * rsum0 % MOD) % MOD;
  return (lans + rans) % MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int Q;
  cin >> N >> Q;
  for (int i = 1; i <= N; i++) {
    cin >> pos[i];
  }
  for (int i = 1; i <= N; i++) {
    cin >> w[i];
    tree.update(i, 0, N, 1);
  }
  for (int i = 0; i < Q; i++) {
    int a, b;
    cin >> a >> b;
    if (a < 0) {
      w[-a] = b;
      tree.update(-a, 0, N, 1);
    } else {
      cout << (query(a, b) + MOD) % MOD << "\n";
    }
  }
}