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>
using namespace std;
const int N = 2e5 + 10;
const int MOD = 1e9 + 7;
inline void add(long long &a, const long long &b) {
  if ((a += b) >= MOD) a -= MOD;
}
int a[N], w[N];
long long c1[N], c2[N];
inline long long sum1(int x) {
  long long ret = 0;
  while (x) ret += c1[x], x -= (x & -x);
  return ret;
}
inline void add1(int x, long long d) {
  while (x < N) c1[x] += d, x += (x & -x);
}
inline long long sum2(int x) {
  long long ret = 0;
  while (x) add(ret, c2[x]), x -= (x & -x);
  return ret;
}
inline void add2(int x, long long d) {
  while (x < N) add(c2[x], d), x += (x & -x);
}
inline void work() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    a[i] -= i;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", w + i);
    add1(i, w[i]);
    add2(i, 1LL * w[i] * a[i] % MOD);
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      add1(-x, -w[-x] + y);
      add2(-x, (1LL * (-w[-x] + y) * a[-x] % MOD + MOD) % MOD);
      w[-x] = y;
    } else {
      int L = x, R = y;
      long long t = sum1(L - 1), sum = sum1(R) - t;
      while (L < R) {
        int M = (L + R) >> 1;
        if (2 * (sum1(M) - t) < sum)
          L = M + 1;
        else
          R = M;
      }
      long long t1 =
          ((sum1(L) - t) % MOD * a[L] % MOD - (sum2(L) - sum2(x - 1)) % MOD) %
          MOD;
      long long t2 = (-(sum1(y) - sum1(L)) % MOD * a[L] % MOD +
                      (sum2(y) - sum2(L)) % MOD) %
                     MOD;
      long long ans = (t1 + t2) % MOD;
      printf("%lld\n", (ans + MOD) % MOD);
    }
  }
}
int main() {
  work();
  return 0;
}
