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]. 
#include <bits/stdc++.h>
int t1, t2, t3, t4;
int n, q;
int q1, q2;
long long pos[262144];
int w[262144];
int seg1[524288];
int seg2[524288];
long long segw[524288];
long long segc1[524288];
long long segc2[524288];
long long getsegw(int u) {
  if (u < 0) return 0;
  long long out = 0;
  u += 262144;
  while (u & (u + 1)) {
    if (u & 1)
      u /= 2;
    else {
      out += segw[u];
      u = u / 2 - 1;
    }
  }
  return out;
}
int best;
long long ans;
void segmove1(int u) {
  if (seg2[u] < q1 || best < seg1[u]) return;
  if (q1 <= seg1[u] && seg2[u] <= best) {
    ans += segc2[u];
    ans += (segw[u] % 1000000007) * (pos[best] - pos[seg2[u]]);
    ans %= 1000000007;
    return;
  }
  segmove1(u * 2);
  segmove1(u * 2 + 1);
}
void segmove2(int u) {
  if (seg2[u] < best || q2 < seg1[u]) return;
  if (best <= seg1[u] && seg2[u] <= q2) {
    ans += segc1[u];
    ans += (segw[u] % 1000000007) * (pos[seg1[u]] - pos[best]);
    ans %= 1000000007;
    return;
  }
  segmove2(u * 2);
  segmove2(u * 2 + 1);
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", pos + i);
    pos[i] -= i;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", w + i);
  }
  for (int i = 0; i < 262144; i++) {
    seg1[i + 262144] = seg2[i + 262144] = i;
  }
  for (int i = 262144 - 1; i > 0; i--) {
    seg1[i] = seg1[i * 2];
    seg2[i] = seg2[i * 2 + 1];
  }
  for (int i = 0; i < n; i++) {
    t1 = i + 262144;
    t2 = w[i];
    do {
      segw[t1] += t2;
      segc1[t1] += t2 * (pos[i] - pos[seg1[t1]]);
      segc1[t1] %= 1000000007;
      segc2[t1] += t2 * (pos[seg2[t1]] - pos[i]);
      segc2[t1] %= 1000000007;
    } while (t1 /= 2);
  }
  while (q--) {
    scanf("%d%d", &q1, &q2);
    if (q1 < 0) {
      int i = -q1 - 1;
      t1 = 262144 + i;
      t2 = q2 - w[i];
      do {
        segw[t1] += t2;
      } while (t1 /= 2);
      t1 = 262144 + i;
      t2 = (q2 + 1000000007 - w[i]) % 1000000007;
      do {
        segc1[t1] += t2 * (pos[i] - pos[seg1[t1]]);
        segc1[t1] %= 1000000007;
        segc2[t1] += t2 * (pos[seg2[t1]] - pos[i]);
        segc2[t1] %= 1000000007;
      } while (t1 /= 2);
      w[i] = q2;
    } else {
      q1--;
      q2--;
      long long w1 = getsegw(q1 - 1);
      long long w2 = getsegw(q2);
      long long remain = w2 + w1;
      t1 = 1;
      while (t1 < 262144) {
        if (remain > segw[t1 * 2] * 2) {
          remain -= segw[t1 * 2] * 2;
          t1 = t1 * 2 + 1;
        } else
          t1 *= 2;
      }
      best = t1 - 262144;
      ans = 0;
      segmove1(1);
      segmove2(1);
      printf("%I64d\n", (ans % 1000000007 + 1000000007) % 1000000007);
    }
  }
}
