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 = 2e5 + 10;
const int MOD = 1e9 + 7;
long long dist[N], cnt[N], a[N], w[N], L[N], R[N];
void update(long long* f, int p, long long x, int flg = 0) {
  while (p < N) {
    if (flg)
      f[p] += x;
    else {
      if ((f[p] += x) >= MOD) f[p] -= MOD;
    };
    p += p & -p;
  }
}
long long query(long long* f, int p, int flg = 0) {
  long long res = 0;
  while (p) {
    if (flg)
      res += f[p];
    else {
      if ((res += f[p]) >= MOD) res -= MOD;
    };
    p -= p & -p;
  }
  return res;
}
long long sum(long long* f, int l, int r, int flg = 0) {
  if (l > r) return 0;
  if (flg) return query(f, r, 1) - query(f, l - 1, 1);
  long long res = query(f, r);
  {
    if ((res += MOD - query(f, l - 1)) >= MOD) res -= MOD;
  };
  return res;
}
long long qsm(long long* f, int l, int r, long long x, int flg = 0) {
  if (l > r) return 0;
  long long res = sum(f, l, r, flg);
  {
    if ((res += MOD - x * (sum(cnt, l, r, 1) % MOD) % MOD) >= MOD) res -= MOD;
  };
  return res;
}
int search(int ql, int qr) {
  int l = ql, r = qr, res = ql;
  long long tot = sum(cnt, ql, qr, 1) / 2;
  while (l <= r) {
    int mid = l + r >> 1;
    if (sum(cnt, ql, mid, 1) <= tot) {
      res = mid + 1;
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  return min(qr, res);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> w[i];
    update(dist, i, (long long)a[i] * w[i] % MOD);
    update(cnt, i, w[i], 1);
    update(L, i, (n - i + 1LL) * w[i] % MOD);
    update(R, i, (long long)i * w[i] % MOD);
  }
  while (q--) {
    int x, y;
    cin >> x >> y;
    if (x < 0) {
      x = -x;
      update(cnt, x, y - w[x], 1);
      long long ch = (y - w[x] + MOD) % MOD;
      w[x] = y;
      update(dist, x, a[x] * ch % MOD);
      update(L, x, (n - x + 1LL) * ch % MOD);
      update(R, x, x * ch % MOD);
    } else {
      int pos = search(x, y);
      long long cl = sum(cnt, x, pos - 1, 1) % MOD,
                cr = sum(cnt, pos + 1, y, 1) % MOD;
      long long res = 0;
      {
        if ((res += a[pos] * cl % MOD) >= MOD) res -= MOD;
      };
      {
        if ((res += MOD - sum(dist, x, pos - 1)) >= MOD) res -= MOD;
      };
      {
        if ((res += sum(dist, pos + 1, y)) >= MOD) res -= MOD;
      };
      {
        if ((res += MOD - a[pos] * cr % MOD) >= MOD) res -= MOD;
      };
      {
        if ((res += MOD - qsm(L, x, pos - 1, n - pos + 1)) >= MOD) res -= MOD;
      };
      {
        if ((res += MOD - qsm(R, pos + 1, y, pos)) >= MOD) res -= MOD;
      };
      cout << res << endl;
    }
  }
  return 0;
}