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>
#pragma warning(disable : 4996)
using namespace std;
template <class T>
inline void chkmin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline void chkmax(T &a, T b) {
  if (a < b) a = b;
}
const int MOD = 1e9 + 7;
const int MAXN = 1 << 18;
int a[MAXN];
long long w[MAXN + MAXN];
int wa[MAXN + MAXN];
void add(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
int sum(int a, int b) { return a + b >= MOD ? a + b - MOD : a + b; }
int mul(long long a, long long b) { return (a * b) % (long long)MOD; }
void Init() {
  for (int i = MAXN - 1; i; --i) {
    wa[i] = sum(wa[2 * i], wa[2 * i + 1]);
    w[i] = w[2 * i] + w[2 * i + 1];
  }
}
void ChWA(int v, int d) {
  wa[v] = d;
  for (v /= 2; v; v /= 2) wa[v] = sum(wa[2 * v], wa[2 * v + 1]);
}
int GetWA(int v, int l, int r, int L, int R) {
  if (r <= L || R <= l) return 0;
  if (L <= l && r <= R) return wa[v];
  int m = (l + r) / 2;
  return sum(GetWA(2 * v, l, m, L, R), GetWA(2 * v + 1, m, r, L, R));
}
void ChW(int v, long long d) {
  w[v] = d;
  for (v /= 2; v; v /= 2) w[v] = w[2 * v] + w[2 * v + 1];
}
long long GetW(int v, int l, int r, int L, int R) {
  if (r <= L || R <= l) return 0;
  if (L <= l && r <= R) return w[v];
  int m = (l + r) / 2;
  return GetW(2 * v, l, m, L, R) + GetW(2 * v + 1, m, r, L, R);
}
int CalcRes(int i, int L, int R) {
  int res = mul(a[i], GetW(1, 0, MAXN, L, i) % (long long)MOD);
  add(res, -GetWA(1, 0, MAXN, L, i) + MOD);
  add(res, GetWA(1, 0, MAXN, i + 1, R));
  add(res, -mul(a[i], GetW(1, 0, MAXN, i + 1, R) % (long long)MOD) + MOD);
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    a[i] -= i;
  }
  for (int i = 0; i < n; ++i) {
    cin >> w[MAXN + i];
    wa[MAXN + i] = mul(a[i], w[MAXN + i]);
  }
  Init();
  for (int qq = 0; qq < q; ++qq) {
    int L;
    cin >> L;
    if (L > 0) {
      --L;
      int R;
      cin >> R;
      int l = L, r = R;
      while (r - l > 1) {
        int m = (l + r) / 2;
        if (GetW(1, 0, MAXN, L, m + 1) <= GetW(1, 0, MAXN, m + 1, R))
          l = m;
        else
          r = m;
      }
      if (GetW(1, 0, MAXN, L, r) < GetW(1, 0, MAXN, r, R))
        cout << CalcRes(r, L, R) << '\n';
      else
        cout << CalcRes(l, L, R) << '\n';
    } else {
      L = -L;
      --L;
      int nw;
      cin >> nw;
      ChW(MAXN + L, nw);
      ChWA(MAXN + L, mul(nw, a[L]));
    }
  }
}