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 = 200010;
const int MOD = 1E9 + 7;
int Add(int a, int b, int m = MOD) {
  a += b;
  if (a >= m) a -= m;
  return a;
}
int Sub(int a, int b, int m = MOD) {
  a -= b;
  if (a < 0) a += m;
  return a;
}
int AddTo(int& to, int w, int m = MOD) {
  to += w;
  if (to >= m) to -= m;
  return to;
}
int SubTo(int& to, int w, int m = MOD) {
  to -= w;
  if (to < 0) to += m;
  return to;
}
int Mult(int a, int b, int m = MOD) { return (a * 1LL * b) % m; }
struct Bit {
  vector<int> sum;
  Bit() { sum.resize(MAXN, 0); }
  void BitAdd(int pos, int w) {
    for (++pos; pos < MAXN; pos += (pos & (-pos))) AddTo(sum[pos], w);
  }
  void BitSub(int pos, int w) {
    for (++pos; pos < MAXN; pos += (pos & (-pos))) SubTo(sum[pos], w);
  }
  int BitGet(int l, int r) {
    int res = 0;
    for (++r; r > 0; r -= (r & (-r))) AddTo(res, sum[r]);
    for (; l > 0; l -= (l & (-l))) SubTo(res, sum[l]);
    return res;
  }
};
struct BitReg {
  vector<long long> sum;
  BitReg() { sum.resize(MAXN, 0); }
  void BitAdd(int pos, int w) {
    for (++pos; pos < MAXN; pos += (pos & (-pos))) sum[pos] += w;
  }
  void BitSub(int pos, int w) {
    for (++pos; pos < MAXN; pos += (pos & (-pos))) sum[pos] -= w;
  }
  long long BitGet(int l, int r) {
    long long res = 0;
    for (++r; r > 0; r -= (r & (-r))) res += sum[r];
    for (; l > 0; l -= (l & (-l))) res -= sum[l];
    return res;
  }
};
int n, q;
int a_[MAXN], w_[MAXN];
int a[MAXN], w[MAXN];
Bit lr, rl;
BitReg sumw;
void Update(int i, int na, int nw) {
  sumw.BitSub(i, w[i]);
  lr.BitSub(i, Mult(Sub(i, a[i]), w[i]));
  rl.BitSub(i, Mult(Sub(a[i], i), w[i]));
  a[i] = na;
  w[i] = nw;
  sumw.BitAdd(i, w[i]);
  lr.BitAdd(i, Mult(Sub(i, a[i]), w[i]));
  rl.BitAdd(i, Mult(Sub(a[i], i), w[i]));
}
int Eval(int l, int r, int mid, int start) {
  int ans = 0;
  if (l <= mid) {
    int koef = Sub(start, l);
    AddTo(ans, lr.BitGet(l, mid));
    AddTo(ans, Mult(koef, sumw.BitGet(l, mid) % MOD));
  }
  if (mid + 1 <= r) {
    int pos = start + mid + 1 - l;
    int koef = Sub(mid + 1, pos);
    AddTo(ans, rl.BitGet(mid + 1, r));
    AddTo(ans, Mult(koef, sumw.BitGet(mid + 1, r) % MOD));
  }
  return ans;
}
int GetMid(int l, int r, int start) {
  if (a[l] > start) return l - 1;
  int lo = l, hi = r, mid;
  for (;;) {
    if (lo + 1 >= hi) {
      if (a[hi] <= start + hi - l)
        return hi;
      else
        return lo;
    }
    mid = (lo + hi) >> 1;
    if (a[mid] <= start + mid - l)
      lo = mid;
    else
      hi = mid;
  }
}
int EvalAtStart(int l, int r, int start) {
  int mid = GetMid(l, r, start);
  return Eval(l, r, mid, start);
}
int Solve(int l, int r) {
  long long total = sumw.BitGet(l, r);
  int lo = l, hi = r, mid;
  for (;;) {
    if (lo + 1 >= hi) {
      if (2 * sumw.BitGet(l, lo) >= total)
        mid = lo;
      else
        mid = hi;
      break;
    }
    mid = (lo + hi) >> 1;
    if (2 * sumw.BitGet(l, mid) <= total)
      lo = mid;
    else
      hi = mid;
  }
  return EvalAtStart(l, r, a[mid] - (mid - l));
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; ++i) scanf("%d", &a_[i]);
  for (int i = 0; i < n; ++i) scanf("%d", &w_[i]);
  for (int i = 0; i < n; ++i) Update(i, a_[i], w_[i]);
  for (int i = 0; i < q; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x > 0) {
      --x, --y;
      printf("%d\n", Solve(x, y));
    } else {
      x = -x - 1;
      Update(x, a[x], y);
    }
  }
  return 0;
}