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;
long long X[200200];
int Xi[200200], xi;
int X2[200200], x2;
int Xm[200200], xm;
int a[200200], w[200200], x, n, q;
inline void update(int u) {
  for (; u <= n; u += u & -u) {
    X[u] += x;
    Xi[u] = (Xi[u] + xi) % 1000000007;
    X2[u] = (X2[u] + x2) % 1000000007;
    Xm[u] = (Xm[u] + xm) % 1000000007;
  }
}
inline long long calc_x(int u, long long ans = 0) {
  for (; u; u -= u & -u) ans += X[u];
  return ans;
}
inline void calc(int u) {
  x = 0, xi = 0, x2 = 0, xm = 0;
  for (; u; u -= u & -u) {
    x = (x + X[u]) % 1000000007;
    xi = (xi + Xi[u]) % 1000000007;
    x2 = (x2 + X2[u]) % 1000000007;
    xm = (xm + Xm[u]) % 1000000007;
  }
}
int collect(int u, int L, int R) {
  calc(L - 1);
  int tx = x, txi = xi, tx2 = x2, txm = xm;
  calc(u - 1);
  tx = (x - tx + 1000000007) % 1000000007,
  txi = (xi - txi + 1000000007) % 1000000007,
  tx2 = (x2 - tx2 + 1000000007) % 1000000007,
  txm = (xm - txm + 1000000007) % 1000000007;
  long long ans = 0;
  if (L < u) {
    ans += (long long)a[u] * tx;
    ans -= txm;
    ans += txi - (long long)u * tx;
  }
  tx = (x + w[u]) % 1000000007, txi = (xi + (long long)u * w[u]) % 1000000007,
  tx2 = (x2 + (long long)w[u] * w[u]) % 1000000007,
  txm = (xm + (long long)a[u] * w[u]) % 1000000007;
  calc(R);
  tx = (x - tx + 1000000007) % 1000000007,
  txi = (xi - txi + 1000000007) % 1000000007,
  tx2 = (x2 - tx2 + 1000000007) % 1000000007,
  txm = (xm - txm + 1000000007) % 1000000007;
  if (u < R) {
    ans += txm;
    ans -= (long long)a[u] * tx;
    ans += (long long)u * tx - txi;
  }
  ans %= 1000000007;
  if (ans < 0) ans += 1000000007;
  return ans;
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  for (int i = 1; i <= n; i++) {
    scanf("%d", w + i);
    x = w[i];
    xi = (long long)i * x % 1000000007;
    x2 = (long long)x * x % 1000000007;
    xm = (long long)a[i] * x % 1000000007;
    update(i);
  }
  while (q--) {
    int L, R;
    scanf("%d %d", &L, &R);
    if (L < 0) {
      L = -L;
      x = -w[L];
      xi = 1000000007 - (long long)L * w[L] % 1000000007;
      x2 = 1000000007 - (long long)w[L] * w[L] % 1000000007;
      xm = 1000000007 - (long long)a[L] * w[L] % 1000000007;
      update(L);
      w[L] = R;
      x = w[L];
      xi = (long long)L * x % 1000000007;
      x2 = (long long)x * x % 1000000007;
      xm = (long long)a[L] * x % 1000000007;
      update(L);
    } else if (L == R)
      puts("0");
    else {
      long long L_ = calc_x(L - 1), all = calc_x(R) - L_;
      int st = L - 1, ed = R;
      while (st + 1 < ed) {
        int md = st + ed >> 1;
        long long sa = calc_x(md) - L_, sb = all - sa;
        if (sa > sb)
          ed = md;
        else
          st = md;
      }
      printf("%d\n", collect(ed, L, R));
    }
  }
}