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 = 212345;
const int MOD = 1e9 + 7;
const int L = 1;
const int R = 1e9;
struct Tree {
  void Clear(int n, long long *base, bool mod) {
    n_ = n;
    mod_ = mod;
    memset(s_, 0, sizeof(s_));
    Build(base, 1, n, 1);
  }
  void Update(int rt) {
    s_[rt] = s_[rt << 1] + s_[rt << 1 | 1];
    if (mod_ && s_[rt] >= MOD) {
      s_[rt] -= MOD;
    }
  }
  void Build(long long *base, int l, int r, int rt) {
    if (l == r) {
      s_[rt] = base[l];
      assert(base[l] < MOD);
      return;
    }
    int m = l + r >> 1;
    Build(base, l, m, rt << 1);
    Build(base, m + 1, r, rt << 1 | 1);
    Update(rt);
  }
  long long RangeSum(int l, int r) { return RangeSum(l, r, 1, n_, 1); }
  long long RangeSum(int xl, int xr, int l, int r, int rt) {
    if (xl <= l && r <= xr) {
      return s_[rt];
    }
    long long sum = 0;
    int m = l + r >> 1;
    if (xl <= m) {
      sum += RangeSum(xl, xr, l, m, rt << 1);
    }
    if (m < xr) {
      sum += RangeSum(xl, xr, m + 1, r, rt << 1 | 1);
    }
    if (mod_ && sum >= MOD) {
      sum -= MOD;
    }
    return sum;
  }
  void SetValue(int x, long long v) { SetValue(x, v, 1, n_, 1); }
  void SetValue(int x, long long v, int l, int r, int rt) {
    if (l == r) {
      s_[rt] = v;
      return;
    }
    int m = l + r >> 1;
    if (x <= m) {
      SetValue(x, v, l, m, rt << 1);
    } else {
      SetValue(x, v, m + 1, r, rt << 1 | 1);
    }
    Update(rt);
  }
  int n_;
  long long s_[N << 2];
  bool mod_;
} sum_left, sum_right, sum_ori;
long long a[N], w[N];
long long wl[N], wr[N];
int Anchor(const int ml, const int mr, const long long sum) {
  int l = ml, r = mr;
  while (l != r) {
    int m = l + r >> 1;
    long long msum = sum_ori.RangeSum(ml, m);
    if (msum * 2 >= sum) {
      r = m;
    } else {
      l = m + 1;
    }
  }
  return l;
}
int GetLeft(int p, int r) {
  if (p >= r) {
    return 0;
  }
  long long lsum = sum_left.RangeSum(p + 1, r);
  long long dist = a[p] - (L + p - 1);
  long long sum = sum_ori.RangeSum(p + 1, r) % MOD;
  return (lsum - dist * sum % MOD + MOD) % MOD;
}
int GetRight(int l, int p, int n) {
  if (l >= p) {
    return 0;
  }
  long long rsum = sum_right.RangeSum(l, p - 1);
  long long dist = R - (n - p) - a[p];
  long long sum = sum_ori.RangeSum(l, p - 1) % MOD;
  return (rsum - dist * sum % MOD + MOD) % MOD;
}
int main() {
  int n, q;
  while (~scanf("%d%d", &n, &q)) {
    for (int i = 1; i <= n; ++i) scanf("%I64d", a + i);
    for (int i = 1; i <= n; ++i) scanf("%I64d", w + i);
    sum_ori.Clear(n, w, false);
    for (int i = 1; i <= n; ++i) {
      wl[i] = w[i] * (a[i] - (L + i - 1)) % MOD;
      wr[i] = w[i] * ((R - (n - i)) - a[i]) % MOD;
    }
    sum_left.Clear(n, wl, true);
    sum_right.Clear(n, wr, true);
    for (int i = 0; i < q; ++i) {
      long long l, r;
      scanf("%I64d%I64d", &l, &r);
      if (l < 0) {
        int x = -l;
        w[x] = r;
        sum_ori.SetValue(x, w[x]);
        wl[x] = w[x] * (a[x] - (L + x - 1)) % MOD;
        sum_left.SetValue(x, wl[x]);
        wr[x] = w[x] * ((R - (n - x)) - a[x]) % MOD;
        sum_right.SetValue(x, wr[x]);
      } else if (l > 0) {
        int len = r - l + 1;
        long long sum = sum_ori.RangeSum(l, r);
        int pivot = Anchor(l, r, sum);
        int ans = 0;
        ans += GetRight(l, pivot, n);
        ans += GetLeft(pivot, r);
        printf("%d\n", ans % MOD);
      } else {
        assert(false);
      }
    }
  }
  return 0;
}