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 maxi = 1e6 + 2;
long long a[maxi], w[maxi];
int n, q;
long long mo = 1e9 + 7;
long long bit[maxi];
long long bit1[maxi];
long long bit2[maxi];
void update(long long *bit, int x, long long val) {
  for (int i = x; i < maxi; i += i & (-i)) bit[i] += val;
  return;
}
long long get(long long *bit, int x) {
  long long ans = 0;
  for (int i = x; i > 0; i -= i & (-i)) ans += bit[i];
  return ans;
}
void update2(long long *bit2, int x, long long val) {
  for (int i = x; i < maxi; i += i & (-i)) {
    bit2[i] = (bit2[i] + val) % mo;
    if (bit2[i] < 0) bit2[i] += mo;
  }
  return;
}
long long get2(long long *bit2, int x) {
  long long ans = 0;
  for (int i = x; i > 0; i -= i & (-i)) {
    ans += bit2[i];
    ans %= mo;
  }
  return ans;
}
long long sredi(long long x) {
  x %= mo;
  if (x < 0) x += mo;
  return x;
}
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%lld", &w[i]);
  for (int i = 1; i <= n; i++) update2(bit2, i, w[i] * a[i]);
  for (int i = 1; i <= n; i++) update(bit, i, w[i]);
  for (int i = 1; i <= n; i++) update2(bit1, i, 1ll * i * w[i]);
  while (q--) {
    int tip;
    int idx, l;
    scanf("%d", &tip);
    if (tip < 0) {
      idx = -tip;
      tip = 1;
    } else {
      l = tip;
      tip = 2;
    }
    if (tip == 1) {
      long long val;
      scanf("%lld", &val);
      update(bit, idx, -w[idx]);
      update2(bit1, idx, -1ll * idx * w[idx]);
      update2(bit2, idx, -1ll * a[idx] * w[idx]);
      w[idx] = val;
      update(bit, idx, w[idx]);
      update2(bit1, idx, 1ll * idx * w[idx]);
      update2(bit2, idx, 1ll * a[idx] * w[idx]);
    } else {
      int r;
      scanf("%d", &r);
      if (l == r)
        printf("0\n");
      else {
        int li = -1;
        int ri = r - l + 1;
        long long sum = get(bit, r) - get(bit, l - 1);
        long long pola = (sum + 1) / 2;
        while (li < ri - 1) {
          int mid = li + ri >> 1;
          long long val = get(bit, l + mid) - get(bit, l - 1);
          if (val >= pola)
            ri = mid;
          else
            li = mid;
        }
        int poz = l + ri;
        long long ans = 0;
        if (poz - 1 >= l) {
          long long x = a[poz] - poz;
          long long sumWi = get(bit, poz - 1) - get(bit, l - 1);
          sumWi %= mo;
          ans += sumWi * x;
          ans = sredi(ans);
          ans += get2(bit1, poz - 1) - get2(bit1, l - 1);
          ans = sredi(ans);
          if (ans < 0) ans += mo;
          ans -= (get2(bit2, poz - 1) - get2(bit2, l - 1));
          ans = sredi(ans);
        }
        if (poz + 1 <= r) {
          long long x = a[poz] - poz;
          ans += get2(bit2, r) - get2(bit2, poz);
          ans = sredi(ans);
          ans -= get2(bit1, r) - get2(bit1, poz);
          ans = sredi(ans);
          long long sumWi = get(bit, r) - get(bit, poz);
          sumWi %= mo;
          ans -= sumWi * x;
          ans = sredi(ans);
        }
        printf("%lld\n", ans);
      }
    }
  }
  return 0;
}
