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 N = (int)2e5 + 5;
const int MOD = (int)1e9 + 7;
const int inf = 0x3f3f3f3f;
struct Seg {
  long long w;
  int aw, iw;
};
int w[N], a[N];
Seg seg[N << 2];
inline int addMod(int a, int b) { return a + b >= MOD ? a + b - MOD : a + b; }
inline Seg merge(const Seg& a, const Seg& b) {
  Seg ret;
  ret.aw = addMod(a.aw, b.aw);
  ret.w = a.w + b.w;
  ret.iw = addMod(a.iw, b.iw);
  return ret;
}
inline void update(int pos, int l, int r, int rt) {
  if (l == r) {
    seg[rt].aw = (long long)a[pos] * w[pos] % MOD;
    seg[rt].w = w[pos];
    seg[rt].iw = (long long)pos * w[pos] % MOD;
    return;
  }
  int m = (l + r) >> 1;
  if (pos <= m) {
    update(pos, l, m, rt << 1);
  } else {
    update(pos, m + 1, r, rt << 1 | 1);
  }
  seg[rt] = merge(seg[rt << 1], seg[rt << 1 | 1]);
}
inline int findK(long long w, int l, int r, int rt) {
  if (l == r) {
    return l;
  }
  int m = (l + r) >> 1;
  if (seg[rt << 1].w > w) {
    return findK(w, l, m, rt << 1);
  } else {
    return findK(w - seg[rt << 1].w, m + 1, r, rt << 1 | 1);
  }
}
inline Seg query(int ql, int qr, int l, int r, int rt) {
  if (ql <= l && r <= qr) {
    return seg[rt];
  }
  int m = (l + r) >> 1;
  Seg ret = Seg{0, 0, 0};
  if (ql <= m) {
    ret = merge(ret, query(ql, qr, l, m, rt << 1));
  }
  if (m < qr) {
    ret = merge(ret, query(ql, qr, m + 1, r, rt << 1 | 1));
  }
  return ret;
}
int main() {
  int n, q;
  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]);
    update(i, 1, n, 1);
  }
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      x = -x;
      w[x] = y;
      update(x, 1, n, 1);
    } else {
      long long pref = x == 1 ? 0 : query(1, x - 1, 1, n, 1).w;
      pref = pref + query(x, y, 1, n, 1).w / 2;
      int k = findK(pref, 1, n, 1);
      Seg t1 = query(x, k, 1, n, 1);
      Seg t2 = k == y ? Seg{0, 0, 0} : query(k + 1, y, 1, n, 1);
      int ans = addMod(t2.aw, MOD - t1.aw);
      ans = addMod(
          ans, (long long)a[k] * addMod(t1.w % MOD, MOD - t2.w % MOD) % MOD);
      ans = addMod(ans,
                   (long long)k * addMod(t2.w % MOD, MOD - t1.w % MOD) % MOD);
      ans = addMod(ans, addMod(t1.iw, MOD - t2.iw));
      printf("%d\n", ans);
    }
  }
}