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 + 5;
const long long MOD = 1e9 + 7;
int n, m;
struct Box {
  long long loc, weight;
} a[N];
struct Push_R {
  long long l, r, weight, cost;
};
Push_R operator+(const Push_R a, const Push_R b) {
  Push_R res;
  res.weight = (a.weight + b.weight) % MOD;
  res.l = b.l - (a.r - a.l + 1), res.r = b.r;
  res.cost = (a.cost + b.cost + a.weight * (b.l - a.r - 1)) % MOD;
  return res;
}
struct Push_L {
  long long l, r, weight, cost;
};
Push_L operator+(const Push_L a, const Push_L b) {
  Push_L res;
  res.weight = (a.weight + b.weight) % MOD;
  res.l = a.l, res.r = a.r + (b.r - b.l + 1);
  res.cost = (a.cost + b.cost + b.weight * (b.l - a.r - 1)) % MOD;
  return res;
}
struct Seg_Tree {
  struct Node {
    Push_L L;
    Push_R R;
  } tr[N << 2];
  void up(int u) {
    tr[u].L = tr[u << 1].L + tr[u << 1 | 1].L;
    tr[u].R = tr[u << 1].R + tr[u << 1 | 1].R;
  }
  void build(int u, int l, int r, Box a[]) {
    if (l == r) {
      tr[u].L.l = tr[u].L.r = a[l].loc;
      tr[u].L.weight = a[l].weight, tr[u].L.cost = 0;
      tr[u].R.l = tr[u].R.r = a[l].loc;
      tr[u].R.weight = a[l].weight, tr[u].R.cost = 0;
      return;
    }
    int mid = (l + r) >> 1;
    build(u << 1, l, mid, a);
    build(u << 1 | 1, mid + 1, r, a);
    up(u);
  }
  void chg(int u, int l, int r, int x, Box z) {
    if (l == r) {
      tr[u].L.weight = z.weight;
      tr[u].R.weight = z.weight;
      return;
    }
    int mid = (l + r) >> 1;
    if (x <= mid)
      chg(u << 1, l, mid, x, z);
    else
      chg(u << 1 | 1, mid + 1, r, x, z);
    up(u);
  }
  Push_L query_L(int u, int l, int r, int x, int y) {
    if (x <= l && r <= y) return tr[u].L;
    int mid = (l + r) >> 1;
    Push_L tmp1, tmp2;
    if (x <= mid) tmp1 = query_L(u << 1, l, mid, x, y);
    if (y > mid) tmp2 = query_L(u << 1 | 1, mid + 1, r, x, y);
    if (x <= mid && y > mid) return tmp1 + tmp2;
    return x <= mid ? tmp1 : tmp2;
  }
  Push_R query_R(int u, int l, int r, int x, int y) {
    if (x <= l && r <= y) return tr[u].R;
    int mid = (l + r) >> 1;
    Push_R tmp1, tmp2;
    if (x <= mid) tmp1 = query_R(u << 1, l, mid, x, y);
    if (y > mid) tmp2 = query_R(u << 1 | 1, mid + 1, r, x, y);
    if (x <= mid && y > mid) return tmp1 + tmp2;
    return x <= mid ? tmp1 : tmp2;
  }
} t1;
struct Tree_Array {
  long long tr[N];
  int lowbit(int x) { return x & (-x); }
  void add(int k, long long x) {
    for (; k <= n; k += lowbit(k)) tr[k] += x;
  }
  long long sum(int k) {
    long long res = 0;
    for (; k; k -= lowbit(k)) res += tr[k];
    return res;
  }
} t2;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i].loc);
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i].weight);
  for (int i = 1; i <= n; ++i) t2.add(i, a[i].weight);
  t1.build(1, 1, n, a);
  while (m--) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      t2.add(-x, -a[-x].weight);
      a[-x].weight = y;
      t1.chg(1, 1, n, -x, a[-x]);
      t2.add(-x, a[-x].weight);
    } else {
      int l = x, r = y, tmp = -1;
      while (l <= r) {
        int mid = (l + r) >> 1;
        long long sum1 = t2.sum(mid) - t2.sum(x - 1);
        long long sum2 = t2.sum(y) - t2.sum(mid);
        if (sum1 >= sum2)
          r = mid - 1, tmp = mid;
        else
          l = mid + 1;
      }
      Push_R tmp1 = t1.query_R(1, 1, n, x, tmp);
      Push_L tmp2 = t1.query_L(1, 1, n, tmp, y);
      printf("%lld\n", (tmp1.cost + tmp2.cost) % MOD);
    }
  }
  return 0;
}