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;
template <typename T>
inline T read() {
  T s = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    s = (s << 3) + (s << 1) + ch - 48;
    ch = getchar();
  }
  return s * f;
}
template <typename T>
inline void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
const int N = 2e5 + 100;
const int mod = 1e9 + 7;
long long w[N], a[N];
struct Segment {
  static const int MAX = 2e5 + 100;
  long long sum[MAX << 2];
  void pushup(int node, int l, int r) {
    sum[node] = sum[node << 1] + sum[node << 1 | 1];
  }
  void build(int node, int l, int r) {
    if (l == r) {
      sum[node] = w[l];
      return;
    }
    int mid = l + r >> 1;
    build(node << 1, l, mid);
    build(node << 1 | 1, mid + 1, r);
    pushup(node, l, r);
  }
  void change(int node, int l, int r, int idx, int val) {
    if (l == r) {
      sum[node] = val;
      return;
    }
    int mid = l + r >> 1;
    if (idx <= mid)
      change(node << 1, l, mid, idx, val);
    else
      change(node << 1 | 1, mid + 1, r, idx, val);
    pushup(node, l, r);
  }
  int queryid(int node, int l, int r, int L, int R, long long val) {
    if (l == r) return l;
    int mid = l + r >> 1;
    if (R <= mid)
      return queryid(node << 1, l, mid, L, R, val);
    else if (L > mid)
      return queryid(node << 1 | 1, mid + 1, r, L, R, val);
    else {
      long long lsum = querysum(node << 1, l, mid, L, mid);
      if (lsum >= val)
        return queryid(node << 1, l, mid, L, mid, val);
      else
        return queryid(node << 1 | 1, mid + 1, r, mid + 1, R, val - lsum);
    }
  }
  long long querysum(int node, int l, int r, int L, int R) {
    if (L <= l && R >= r) {
      return sum[node];
    }
    int mid = l + r >> 1;
    long long val = 0;
    if (L <= mid) val += querysum(node << 1, l, mid, L, R);
    if (R > mid) val += querysum(node << 1 | 1, mid + 1, r, L, R);
    return val;
  }
} treeans;
struct SegmentTree {
  static const int MAX = 2e5 + 100;
  long long sum[MAX << 2];
  void pushup(int node, int l, int r) {
    sum[node] = sum[node << 1] + sum[node << 1 | 1] % mod;
  }
  void build(int node, int l, int r) {
    if (l == r) {
      sum[node] = w[l] * (a[l] - l) % mod;
      return;
    }
    int mid = l + r >> 1;
    build(node << 1, l, mid);
    build(node << 1 | 1, mid + 1, r);
    pushup(node, l, r);
  }
  void change(int node, int l, int r, int idx, long long val) {
    if (l == r) {
      sum[node] = val * (a[l] - l) % mod;
      return;
    }
    int mid = l + r >> 1;
    if (idx <= mid)
      change(node << 1, l, mid, idx, val);
    else
      change(node << 1 | 1, mid + 1, r, idx, val);
    pushup(node, l, r);
  }
  long long querysum(int node, int l, int r, int L, int R) {
    if (L <= l && R >= r) {
      return sum[node] % mod;
    }
    int mid = l + r >> 1;
    long long val = 0;
    if (L <= mid) val += querysum(node << 1, l, mid, L, R), val %= mod;
    if (R > mid) val += querysum(node << 1 | 1, mid + 1, r, L, R), val %= mod;
    return val % mod;
  }
} tree;
int main() {
  int n = read<int>(), m = read<int>();
  for (int i = 1; i <= n; ++i) {
    a[i] = read<long long>();
  }
  for (int i = 1; i <= n; ++i) {
    w[i] = read<long long>();
  }
  tree.build(1, 1, n);
  treeans.build(1, 1, n);
  for (int i = 1; i <= m; ++i) {
    long long l = read<long long>(), r = read<long long>();
    if (l < 0) {
      l = -l;
      tree.change(1, 1, n, l, r);
      treeans.change(1, 1, n, l, r);
      w[l] = r;
    } else {
      long long k = treeans.querysum(1, 1, n, l, r);
      if (k & 1) {
        k = (k + 1) / 2;
        long long id = treeans.queryid(1, 1, n, l, r, k);
        long long ans = 0;
        ans = ((1LL * ((a[id] - id) % mod) *
                (treeans.querysum(1, 1, n, l, id) % mod)) %
                   mod -
               1LL * tree.querysum(1, 1, n, l, id) % mod + mod) %
              mod;
        ans = (ans + 1LL * tree.querysum(1, 1, n, id, r) % mod -
               1LL *
                   ((a[id] - id) % mod *
                    (treeans.querysum(1, 1, n, id, r) % mod)) %
                   mod +
               mod) %
              mod;
        printf("%lld\n", (long long)ans % mod);
      } else {
        k /= 2;
        int idone = treeans.queryid(1, 1, n, l, r, k);
        int idtwo = treeans.queryid(1, 1, n, l, r, k + 1);
        long long valmidlf = (a[idone] + a[idtwo]) / 2;
        long long valmidrt = valmidlf;
        if (idone != idtwo) valmidrt += 1;
        long long ans = 0;
        ans = (1LL * (valmidlf - idone) % mod *
                   (treeans.querysum(1, 1, n, l, idone) % mod) % mod -
               1LL * tree.querysum(1, 1, n, l, idone) % mod + mod) %
              mod;
        ans = (ans + 1LL * tree.querysum(1, 1, n, idtwo, r) % mod -
               1LL * (valmidrt - idtwo) % mod *
                   (treeans.querysum(1, 1, n, idtwo, r) % mod) % mod +
               mod) %
              mod;
        printf("%lld\n", (long long)ans % mod);
      }
    }
  }
}