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 long long MAXN = 2e5 + 7;
const long long MOD = 1000 * 1000 * 1000 + 7;
long long mod(long long n) { return (n % MOD + MOD) % MOD; }
long long n, q;
long long a[MAXN], w[MAXN];
void read() {
  cin >> n >> q;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < n; ++i) cin >> w[i];
}
long long f[MAXN], f1[MAXN];
void add_(long long i, long long x) {
  for (; i < MAXN; i |= (i + 1)) {
    f[i] += x;
  }
}
long long getpr_(long long i) {
  long long ans = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) {
    ans += f[i];
  }
  return ans;
}
long long get_(long long l, long long r) { return getpr_(r) - getpr_(l - 1); }
void add1_(long long i, long long x) {
  long long t = x * (i - a[i]);
  for (; i < MAXN; i |= (i + 1)) {
    f1[i] += t;
    f1[i] = mod(f1[i]);
  }
}
long long getpr1_(long long i) {
  long long ans = 0;
  for (; i >= 0; i = (i & (i + 1)) - 1) {
    ans += f1[i];
  }
  return ans;
}
long long get1_(long long l, long long r) {
  return mod(getpr1_(r) - getpr1_(l - 1));
}
long long p[MAXN];
void solve() {
  for (long long i = 0; i < n; ++i) add_(i, w[i]);
  for (long long i = 0; i < n; ++i) add1_(i, w[i]);
  for (long long i = 0; i < n; ++i) p[i] = a[i] - i;
}
void upd(long long i, long long x) {
  long long delt = x - w[i];
  add_(i, delt);
  add1_(i, delt);
  w[i] = x;
}
long long getl(long long l, long long r) {
  long long bl = a[l] - 1;
  long long br = a[r] - (r - l) + 1;
  long long sum = get_(l, r);
  while (bl < br - 1) {
    long long bm = (bl + br) / 2;
    auto t = upper_bound(p + l, p + r + 1, bm - l);
    long long sum1 = get_(t - p, r);
    if (sum < sum1 * 2)
      bl = bm;
    else
      br = bm;
  }
  return br;
}
long long getans(long long l, long long r, long long tl) {
  auto t = upper_bound(p + l, p + r + 1, tl - l);
  long long bl = t - p - 1;
  long long ans = 0;
  ans += mod(mod(get_(l, bl)) * (tl - l)) + get1_(l, bl);
  ans -= mod(mod(get_(bl + 1, r)) * (tl - l)) + get1_(bl + 1, r);
  return mod(ans);
}
long long get(long long l, long long r) {
  long long tl = getl(l, r);
  return getans(l, r, tl);
}
void print() {
  for (long long i = 0; i < q; ++i) {
    long long x, y;
    cin >> x >> y;
    if (x < 0) {
      upd(-x - 1, y);
    } else {
      cout << get(x - 1, y - 1) << '\n';
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  read();
  solve();
  print();
  return 0;
}