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 = 2e5 + 5;
const int mod = 1e9 + 7;
int a[maxn], w[maxn], sm[maxn << 2], sm2[maxn << 2], n;
long long t[maxn << 2];
void pushup(int rt) {
  t[rt] = t[rt << 1] + t[rt << 1 | 1];
  sm[rt] = (sm[rt << 1] + sm[rt << 1 | 1]) % mod;
  sm2[rt] = (sm2[rt << 1] + sm2[rt << 1 | 1]) % mod;
}
void build(int rt, int l, int r) {
  if (l == r) {
    t[rt] = w[l];
    sm[rt] = 1ll * w[l] * (a[n] - a[l] - (n - l)) % mod;
    sm2[rt] = 1ll * w[l] * (a[l] - a[1] - (l - 1)) % mod;
    return;
  }
  int mid = l + r >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  pushup(rt);
}
void cg(int rt, int l, int r, int x, int v) {
  if (l == r) {
    t[rt] = v;
    sm[rt] = 1ll * v * (a[n] - a[l] - (n - l)) % mod;
    sm2[rt] = 1ll * v * (a[l] - a[1] - (l - 1)) % mod;
    return;
  }
  int mid = l + r >> 1;
  if (x <= mid)
    cg(rt << 1, l, mid, x, v);
  else
    cg(rt << 1 | 1, mid + 1, r, x, v);
  pushup(rt);
}
long long qry(int rt, int l, int r, int ql, int qr) {
  if (l >= ql && r <= qr) return t[rt];
  int mid = l + r >> 1;
  long long ans = 0;
  if (ql <= mid) ans += qry(rt << 1, l, mid, ql, qr);
  if (qr >= mid + 1) ans += qry(rt << 1 | 1, mid + 1, r, ql, qr);
  return ans;
}
int qrysm(int rt, int l, int r, int ql, int qr) {
  if (l >= ql && r <= qr) return sm[rt];
  int mid = l + r >> 1;
  int ans = 0;
  if (ql <= mid) ans = (ans + qrysm(rt << 1, l, mid, ql, qr)) % mod;
  if (qr >= mid + 1) ans = (ans + qrysm(rt << 1 | 1, mid + 1, r, ql, qr)) % mod;
  return ans;
}
int qrysm2(int rt, int l, int r, int ql, int qr) {
  if (l >= ql && r <= qr) return sm2[rt];
  int mid = l + r >> 1;
  int ans = 0;
  if (ql <= mid) ans = (ans + qrysm2(rt << 1, l, mid, ql, qr)) % mod;
  if (qr >= mid + 1)
    ans = (ans + qrysm2(rt << 1 | 1, mid + 1, r, ql, qr)) % mod;
  return ans;
}
int fd(int rt, int l, int r, int ql, int qr, long long &val) {
  int mid = l + r >> 1;
  if (l >= ql && r <= qr) {
    if (t[rt] == val) {
      val -= t[rt];
      return r;
    } else if (t[rt] < val) {
      val -= t[rt];
      return r + 1;
    } else if (t[rt] > val) {
      if (l == r) {
        val -= t[rt];
        return l;
      }
      int ans = fd(rt << 1, l, mid, ql, qr, val);
      if (val <= 0) return ans;
      ans = fd(rt << 1 | 1, mid + 1, r, ql, qr, val);
      return ans;
    }
  } else {
    int ans = 0;
    if (ql <= mid) {
      ans = fd(rt << 1, l, mid, ql, qr, val);
      if (val <= 0) return ans;
    }
    if (qr >= mid + 1) {
      ans = fd(rt << 1 | 1, mid + 1, r, ql, qr, val);
      if (val <= 0) return ans;
    }
  }
  return r + 1;
}
int main() {
  int 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]);
  build(1, 1, n);
  while (q--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      x = -x;
      cg(1, 1, n, x, y);
    } else {
      long long sum = qry(1, 1, n, x, y);
      sum = (sum + 1) / 2;
      int id = fd(1, 1, n, x, y, sum);
      int ans = qrysm(1, 1, n, x, id);
      ans = (ans - 1ll * qry(1, 1, n, x, id) % mod * (a[n] - a[id] - (n - id)) %
                       mod) %
            mod;
      ans = (ans + qrysm2(1, 1, n, id, y)) % mod;
      ans = (ans - 1ll * qry(1, 1, n, id, y) % mod * (a[id] - a[1] - (id - 1)) %
                       mod) %
            mod;
      ans = (ans + mod) % mod;
      printf("%d\n", ans);
    }
  }
}