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 + 3;
const int MOD = (int)1e9 + 7;
const int infint = (int)1e9;
const long long inf = (long long)1e15;
long long n, a[MAXN], q, w[MAXN];
long long mul(long long a, long long b) {
  a %= MOD, b %= MOD;
  long long c = 1LL * a * b;
  c %= MOD;
  return c;
}
long long add(long long a, long long b) {
  a %= MOD, b %= MOD;
  long long c = a + b;
  if (c < 0) c += MOD;
  if (c >= MOD) c -= MOD;
  return c;
}
struct segment {
  long long seg[2 * MAXN];
  void build(bool toMOD) {
    for (int i = n - 1; i > 0; i--) {
      if (toMOD == 0)
        seg[i] = seg[i << 1] + seg[i << 1 | 1];
      else
        seg[i] = add(seg[i << 1], seg[i << 1 | 1]);
    }
  }
  void update(int idx, long long v, bool toMOD) {
    for (seg[idx += n] = v; idx > 0; idx >>= 1) {
      if (toMOD == 0)
        seg[idx >> 1] = seg[idx] + seg[idx ^ 1];
      else
        seg[idx >> 1] = add(seg[idx], seg[idx ^ 1]);
    }
  }
  long long get(int l, int r, bool toMOD) {
    long long sum = 0;
    for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
      if (l & 1) {
        if (toMOD == 0)
          sum += seg[l++];
        else
          sum = add(sum, seg[l++]);
      }
      if (r & 1) {
        if (toMOD == 0)
          sum += seg[--r];
        else
          sum = add(sum, seg[--r]);
      }
    }
    return sum;
  }
} sum, wei;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    cin >> w[i];
    sum.seg[i + n] = w[i];
    wei.seg[i + n] = mul(w[i], add(a[i], -i));
  }
  sum.build(0), wei.build(1);
  for (int i = 0; i < q; i++) {
    int type;
    cin >> type;
    if (type < 0) {
      long long idx, val;
      idx = -type;
      idx--;
      cin >> val;
      sum.update(idx, val, 0);
      wei.update(idx, mul(val, add(a[idx], -idx)), 1);
    } else {
      long long l, r;
      l = type;
      l--;
      cin >> r;
      int L = l - 1, R = r - 1;
      while (R - L > 1) {
        int mid = (L + R) >> 1;
        if (sum.get(l, mid + 1, 0) >= sum.get(mid + 1, r, 0))
          R = mid;
        else
          L = mid;
      }
      long long ans = 0;
      ans = add(mul(sum.get(l, R, 1), add(a[R], -R)), -wei.get(l, R, 1));
      ans = add(ans,
                add(wei.get(R, r, 1), -mul(sum.get(R, r, 1), add(a[R], -R))));
      cout << ans << "\n";
    }
  }
}