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 M = 1e9 + 7;
class FenwickTree {
 public:
  FenwickTree(int _n) : n(_n), data(n, 0) {}
  void add(int x, long long v) {
    for (; x < n; x |= x + 1) {
      data[x] += v;
    }
  }
  long long get(int x) {
    long long res = 0;
    for (; x >= 0; x = (x & (x + 1)) - 1) {
      res += data[x];
    }
    return res;
  }

 private:
  int n;
  vector<long long> data;
};
class Solution {
 public:
  Solution(vector<int>& _pos, vector<int>& _cost, int _n)
      : n(_n), pos(_pos), cost(_cost), cft(n + 10), pcft(n + 10) {
    for (int i = 0; i < n; ++i) {
      pos[i] -= i + 1;
      cft.add(i, cost[i]);
      pcft.add(i, mulM(cost[i], pos[i]));
    }
  }
  void update(int x, long long v) {
    cft.add(x, v - cost[x]);
    pcft.add(x, mulM(v - cost[x], pos[x]));
    cost[x] = v;
  }
  long long query(int l, int r) {
    long long res = 0;
    int lo = l, hi = r;
    long long sum = cft.get(r) - cft.get(l - 1);
    while (lo <= hi) {
      int mi = lo + (hi - lo) / 2;
      long long sum1 = cft.get(mi) - cft.get(l - 1);
      if (sum1 * 2LL < sum) {
        lo = mi + 1;
      } else {
        hi = mi - 1;
      }
    }
    res = subM(mulM(pos[lo], subM(cft.get(lo - 1), cft.get(l - 1))),
               subM(pcft.get(lo - 1), pcft.get(l - 1)));
    res = addM(res, subM(subM(pcft.get(r), pcft.get(lo - 1)),
                         mulM(pos[lo], subM(cft.get(r), cft.get(lo - 1)))));
    return res;
  }

 private:
  int n;
  FenwickTree cft;
  FenwickTree pcft;
  vector<int> pos;
  vector<int> cost;
  long long addM(long long a, long long b) {
    a += b;
    a %= M;
    while (a >= M) {
      a -= M;
    }
    while (a < 0) {
      a += M;
    }
    return a;
  }
  long long subM(long long a, long long b) {
    a -= b;
    a %= M;
    while (a >= M) {
      a -= M;
    }
    while (a < 0) {
      a += M;
    }
    return a;
  }
  long long mulM(long long a, long long b) {
    a *= b;
    a %= M;
    while (a >= M) {
      a -= M;
    }
    while (a < 0) {
      a += M;
    }
    return a;
  }
  long long powM(long long x, long long e) {
    long long res = 1;
    while (e > 0) {
      if (e & 0x1) {
        res = mulM(res, x);
      }
      x = mulM(x, x);
      e >>= 1;
    }
    return res;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q;
  cin >> n >> q;
  vector<int> pos(n);
  vector<int> cost(n);
  for (int i = 0; i < n; ++i) {
    cin >> pos[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> cost[i];
  }
  Solution sol(pos, cost, n);
  for (int i = 0, x = 0, y = 0; i < q; ++i) {
    cin >> x >> y;
    if (x < 0) {
      x = -x;
      --x;
      sol.update(x, y);
    } else {
      --x;
      --y;
      cout << sol.query(x, y) << endl;
    }
  }
  return 0;
}