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 + 10, mod = 1e9 + 7;
void add(int& x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
}
void sub(int& x, int y) {
  x -= y;
  if (x < 0) {
    x += mod;
  }
}
int mul(int x, int y) { return (long long)x * y % mod; }
int n, q, a[N], value[N];
long long sum[N << 2];
void build(int l, int r, int o) {
  if (l == r) {
    sum[o] = value[l];
  } else {
    int mid = l + r >> 1;
    build(l, mid, (o << 1));
    build(mid + 1, r, (o << 1 | 1));
    sum[o] = sum[(o << 1)] + sum[(o << 1 | 1)];
  }
}
void modify(int l, int r, int o, int p) {
  if (l == r) {
    sum[o] = value[l];
  } else {
    int mid = l + r >> 1;
    if (p <= mid) {
      modify(l, mid, (o << 1), p);
    } else {
      modify(mid + 1, r, (o << 1 | 1), p);
    }
    sum[o] = sum[(o << 1)] + sum[(o << 1 | 1)];
  }
}
long long query(int l, int r, int o, int ql, int qr) {
  if (ql <= l && r <= qr) {
    return sum[o];
  } else {
    int mid = l + r >> 1;
    long long result = 0;
    if (ql <= mid) {
      result += query(l, mid, (o << 1), ql, qr);
    }
    if (qr > mid) {
      result += query(mid + 1, r, (o << 1 | 1), ql, qr);
    }
    return result;
  }
}
int query(int l, int r, int o, long long x) {
  if (l == r) {
    return l;
  } else {
    int mid = l + r >> 1;
    if (sum[(o << 1)] > x) {
      return query(l, mid, (o << 1), x);
    } else {
      return query(mid + 1, r, (o << 1 | 1), x - sum[(o << 1)]);
    }
  }
}
struct state {
  int a, b, c;
  state() {}
  state(int a, int b, int c) : a(a), b(b), c(c) {}
  state operator+(const state& other) const {
    state result = *this;
    add(result.a, other.a);
    add(result.b, other.b);
    add(result.c, other.c);
    return result;
  }
  state operator-(const state& other) const {
    state result = *this;
    sub(result.a, other.a);
    sub(result.b, other.b);
    sub(result.c, other.c);
    return result;
  }
} c[N];
void modify(int p, state x) {
  for (; p <= n; p += p & -p) {
    c[p] = c[p] + x;
  }
}
state query(int p) {
  state result(0, 0, 0);
  for (; p; p -= p & -p) {
    result = result + c[p];
  }
  return result;
}
state query(int l, int r) { return query(r) - query(l - 1); }
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &value[i]);
    modify(i, state(value[i], mul(value[i], i), mul(value[i], a[i])));
  }
  build(1, n, 1);
  for (int i = 1; i <= q; ++i) {
    int p, x;
    scanf("%d%d", &p, &x);
    if (p < 0) {
      p = -p;
      modify(p, state((mod - value[p]) % mod, (mod - mul(value[p], p)) % mod,
                      (mod - mul(value[p], a[p])) % mod));
      value[p] = x;
      modify(p, state(value[p], mul(value[p], p), mul(value[p], a[p])));
      modify(1, n, 1, p);
    } else {
      long long sumv = query(1, n, 1, p, x) >> 1;
      if (p ^ 1) {
        sumv += query(1, n, 1, 1, p - 1);
      }
      int static_pos = query(1, n, 1, sumv), ans = 0;
      if (p ^ static_pos) {
        int l = p, r = static_pos - 1;
        state result = query(l, r);
        add(ans,
            ((result.b + (long long)result.a * (a[static_pos] - static_pos) -
              result.c) %
                 mod +
             mod) %
                mod);
      }
      if (x ^ static_pos) {
        int l = static_pos + 1, r = x;
        state result = query(l, r);
        add(ans,
            ((result.c - (long long)result.a * (a[static_pos] - static_pos) -
              result.b) %
                 mod +
             mod) %
                mod);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}