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 long long mod = 1e9 + 7;
int a[400020], w[400020];
int tot, ls[400020], rs[400020];
long long sumw[400020], sum[400020];
int n, q, rt;
inline void update(int x) {
  sum[x] = (sum[ls[x]] + sum[rs[x]]) % mod;
  sumw[x] = sumw[ls[x]] + sumw[rs[x]];
}
void build(int &x, int l, int r) {
  x = ++tot;
  if (l == r) {
    sumw[x] = w[l], sum[x] = (long long)a[l] * w[l] % mod;
    return;
  }
  int mid = (l + r) >> 1;
  build(ls[x], l, mid), build(rs[x], mid + 1, r);
  update(x);
}
void modify(int x, int l, int r, int id, int d) {
  if (l == r) {
    sumw[x] = d, sum[x] = (long long)a[l] * d % mod;
    return;
  }
  int mid = (l + r) >> 1;
  if (id <= mid)
    modify(ls[x], l, mid, id, d);
  else
    modify(rs[x], mid + 1, r, id, d);
  update(x);
}
long long queryw(int x, int l, int r, int L, int R) {
  if (L <= l && R >= r) return sumw[x];
  int mid = (l + r) >> 1;
  long long res = 0;
  if (L <= mid) res = queryw(ls[x], l, mid, L, R);
  if (R > mid) res += queryw(rs[x], mid + 1, r, L, R);
  return res;
}
long long querys(int x, int l, int r, int L, int R) {
  if (L <= l && R >= r) return sum[x];
  int mid = (l + r) >> 1;
  long long res = 0;
  if (L <= mid) res = querys(ls[x], l, mid, L, R);
  if (R > mid) res += querys(rs[x], mid + 1, r, L, R);
  return res % mod;
}
int queryid(int x, int l, int r, int L, int R, long long &D) {
  if (l == r) {
    D -= sumw[x];
    return D <= 0 ? l : 0;
  }
  int mid = (l + r) >> 1, res = 0;
  if (L <= l && R >= r) {
    if (D > sumw[x]) {
      D -= sumw[x];
      return 0;
    }
    if (D <= sumw[ls[x]]) return queryid(ls[x], l, mid, L, R, D);
    D -= sumw[ls[x]];
    return queryid(rs[x], mid + 1, r, L, R, D);
  }
  if (L <= mid) res = queryid(ls[x], l, mid, L, R, D);
  if (res || R <= mid) return res;
  return queryid(rs[x], mid + 1, r, L, R, D);
}
int main() {
  scanf("%d %d", &n, &q);
  for (register int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i] = a[i] - i;
  for (register int i = 1; i <= n; i++) scanf("%d", &w[i]);
  build(rt, 1, n);
  while (q--) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (x < 0) {
      modify(rt, 1, n, -x, y);
    } else {
      long long sumw = queryw(rt, 1, n, x, y), tmp = (sumw + 1) / 2;
      int id = queryid(rt, 1, n, x, y, tmp);
      long long w1 = queryw(rt, 1, n, x, id), w2 = sumw - w1;
      long long ans = (((long long)a[id] * w1 - querys(rt, 1, n, x, id) +
                        querys(rt, 1, n, id + 1, y) - (long long)a[id] * w2) %
                           mod +
                       mod) %
                      mod;
      printf("%lld\n", ans);
    }
  }
}