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 int maxn = (int)2e5 + 1, mod = (int)1e9 + 7;
int n, m, a[maxn];
struct Segment {
  long long cnt;
  int sum;
} seg[maxn << 1 | 1];
inline int seg_idx(int L, int R) { return (L + R) | (L < R); }
inline void seg_merge(Segment &rt, Segment &lch, Segment &rch) {
  rt.cnt = lch.cnt + rch.cnt;
  (rt.sum = lch.sum + rch.sum) >= mod && (rt.sum -= mod);
}
void seg_build(int L, int R) {
  int rt = seg_idx(L, R);
  if (L == R) {
    int v;
    scanf("%d", &v);
    seg[rt] = (Segment){(long long)v, (int)((long long)a[L] * v % mod)};
    return;
  }
  int M = (L + R) >> 1, lch = seg_idx(L, M), rch = seg_idx(M + 1, R);
  seg_build(L, M);
  seg_build(M + 1, R);
  seg_merge(seg[rt], seg[lch], seg[rch]);
}
void seg_upd(int L, int R, int x, int v) {
  int rt = seg_idx(L, R);
  if (L == R) {
    seg[rt] = (Segment){(long long)v, (int)((long long)a[L] * v % mod)};
    return;
  }
  int M = (L + R) >> 1, lch = seg_idx(L, M), rch = seg_idx(M + 1, R);
  if (x <= M)
    seg_upd(L, M, x, v);
  else
    seg_upd(M + 1, R, x, v);
  seg_merge(seg[rt], seg[lch], seg[rch]);
}
Segment seg_que(int L, int R, int x) {
  if (R <= x) return seg[seg_idx(L, R)];
  int M = (L + R) >> 1;
  if (x <= M) return seg_que(L, M, x);
  Segment ret = seg_que(M + 1, R, x);
  seg_merge(ret, seg[seg_idx(L, M)], ret);
  return ret;
}
int seg_find(int L, int R, long long v) {
  while (L < R) {
    int M = (L + R) >> 1;
    if (seg[seg_idx(L, M)].cnt >= v) {
      R = M;
    } else {
      v -= seg[seg_idx(L, M)].cnt;
      L = M + 1;
    }
  }
  return L;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    a[i] -= i;
  }
  seg_build(1, n);
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      seg_upd(1, n, -x, y);
    } else {
      Segment lft = x > 1 ? seg_que(1, n, x - 1) : (Segment){0LL, 0};
      Segment rht = seg_que(1, n, y);
      int M = seg_find(1, n, (lft.cnt + rht.cnt + 1) >> 1);
      Segment mid = seg_que(1, n, M);
      int res =
          ((mid.cnt - lft.cnt) % mod * a[M] - (long long)(mid.sum - lft.sum) +
           (long long)(rht.sum - mid.sum) - (rht.cnt - mid.cnt) % mod * a[M]) %
          mod;
      printf("%d\n", res < 0 ? res + mod : res);
    }
  }
  return 0;
}