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;
int n;
long long mod = (long long)1e9 + 7;
vector<long long> a, w;
vector<long long> segTreeW, segTreeAW;
void build(int id, int l, int r) {
  if (l == r) {
    segTreeW[id] = w[l];
    segTreeAW[id] = w[l] * a[l] % mod;
  } else {
    int mid = (l + r) / 2;
    build(id * 2, l, mid);
    build(id * 2 + 1, mid + 1, r);
    segTreeW[id] = segTreeW[id * 2] + segTreeW[id * 2 + 1];
    segTreeAW[id] = (segTreeAW[id * 2] + segTreeAW[id * 2 + 1]) % mod;
  }
}
void build() {
  segTreeW.resize(n * 4 + 2517);
  segTreeAW.resize(n * 4 + 2517);
  build(1, 0, n - 1);
}
long long sumW(int id, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (l == tl && r == tr) return segTreeW[id];
  int mid = (tl + tr) / 2;
  return sumW(id * 2, tl, mid, l, min(r, mid)) +
         sumW(id * 2 + 1, mid + 1, tr, max(l, mid + 1), r);
}
long long sumW(int l, int r) { return sumW(1, 0, n - 1, l, r); }
long long sumAW(int id, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (l == tl && r == tr) return segTreeAW[id];
  int mid = (tl + tr) / 2;
  return (sumAW(id * 2, tl, mid, l, min(r, mid)) +
          sumAW(id * 2 + 1, mid + 1, tr, max(l, mid + 1), r)) %
         mod;
}
long long sumAW(int l, int r) { return sumAW(1, 0, n - 1, l, r); }
void upd(int id, int tl, int tr, int pos, long long nv) {
  if (tl == tr) {
    segTreeW[id] = w[pos] = nv;
    segTreeAW[id] = w[pos] * a[pos] % mod;
  } else {
    int tm = (tl + tr) / 2;
    if (pos <= tm)
      upd(id * 2, tl, tm, pos, nv);
    else
      upd(id * 2 + 1, tm + 1, tr, pos, nv);
    segTreeW[id] = segTreeW[id * 2] + segTreeW[id * 2 + 1];
    segTreeAW[id] = (segTreeAW[id * 2] + segTreeAW[id * 2 + 1]) % mod;
  }
}
void upd(int pos, long long nv) { upd(1, 0, n - 1, pos, nv); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int q;
  cin >> n >> q;
  a.resize(n);
  for (int(i) = 0; (i) < (n); (i)++) cin >> a[i], a[i] -= i;
  w.resize(n);
  for (int(i) = 0; (i) < (n); (i)++) cin >> w[i];
  build();
  for (int l, r; q-- > 0;) {
    cin >> l >> r;
    if (l < 0) {
      upd(-l - 1, r);
      continue;
    }
    long long sW = sumW(--l, --r);
    int lk = l - 1, mk, rk = r;
    while (lk + 1 < rk) {
      mk = (lk + rk) / 2;
      if (sumW(l, mk) * 2 >= sW)
        rk = mk;
      else
        lk = mk;
    }
    cout << (((sumW(l, rk - 1) % mod * a[rk] % mod + mod - sumAW(l, rk - 1)) %
                  mod +
              (sumAW(rk, r) + mod - sumW(rk, r) % mod * a[rk] % mod) % mod) %
             mod)
         << '\n';
  }
  return 0;
}