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;
void qmax(long long &x, long long y) {
  if (x < y) x = y;
}
void qmin(long long &x, long long y) {
  if (x > y) x = y;
}
inline long long read() {
  char s;
  long long k = 0, base = 1;
  while ((s = getchar()) != '-' && s != EOF && !(isdigit(s)))
    ;
  if (s == EOF) exit(0);
  if (s == '-') base = -1, s = getchar();
  while (isdigit(s)) k = k * 10 + (s ^ '0'), s = getchar();
  return k * base;
}
inline void write(long long x) {
  static char cnt, num[15];
  cnt = 0;
  if (!x) {
    putchar('0');
    return;
  }
  for (; x; x /= 10) num[++cnt] = x % 10;
  for (; cnt; putchar(num[cnt--] + 48))
    ;
}
const long long mod = 1e9 + 7;
const long long maxn = 8e5 + 100;
long long a[maxn], w[maxn];
long long n, q;
long long h[maxn << 2];
long long dis[2][maxn << 2];
void Mod(long long &x) { x %= mod; }
void bt(long long x, long long y, long long d) {
  if (x == y) {
    dis[0][d] = w[x] * (a[x] - x) % mod;
    dis[1][d] = w[x] * (a[n] - (n - x) - a[x]) % mod;
    h[d] = w[x];
    return;
  }
  long long mid = (x + y) / 2;
  bt(x, mid, d * 2);
  bt(mid + 1, y, d * 2 + 1);
  h[d] = h[d * 2] + h[d * 2 + 1];
  dis[0][d] = dis[0][d * 2] + dis[0][d * 2 + 1];
  Mod(dis[0][d]);
  dis[1][d] = dis[1][d * 2] + dis[1][d * 2 + 1];
  Mod(dis[1][d]);
}
long long qsum(long long x, long long y, long long d, long long l,
               long long r) {
  if (x == l && y == r) return h[d];
  long long mid = (x + y) / 2;
  if (r <= mid) return qsum(x, mid, d * 2, l, r);
  if (l > mid) return qsum(mid + 1, y, d * 2 + 1, l, r);
  return qsum(x, mid, d * 2, l, mid) + qsum(mid + 1, y, d * 2 + 1, mid + 1, r);
}
long long kth(long long x, long long y, long long d, long long k) {
  if (x == y) return x;
  long long mid = (x + y) / 2;
  if (h[d * 2] >= k) return kth(x, mid, d * 2, k);
  return kth(mid + 1, y, d * 2 + 1, k - h[d * 2]);
}
long long qdis(long long x, long long y, long long d, long long l, long long r,
               long long flag) {
  if (l == x && y == r) return dis[flag][d];
  long long mid = (x + y) / 2;
  if (r <= mid) return qdis(x, mid, d * 2, l, r, flag);
  if (l > mid) return qdis(mid + 1, y, d * 2 + 1, l, r, flag);
  return (qdis(x, mid, d * 2, l, mid, flag) +
          qdis(mid + 1, y, d * 2 + 1, mid + 1, r, flag)) %
         mod;
}
void xg(long long x, long long y, long long d, long long p) {
  if (x == y) {
    dis[0][d] = w[x] * (a[x] - x) % mod;
    dis[1][d] = w[x] * (a[n] - (n - x) - a[x]) % mod;
    h[d] = w[x];
    return;
  }
  long long mid = (x + y) / 2;
  if (p <= mid)
    xg(x, mid, d * 2, p);
  else
    xg(mid + 1, y, d * 2 + 1, p);
  h[d] = h[d * 2] + h[d * 2 + 1];
  dis[0][d] = dis[0][d * 2] + dis[0][d * 2 + 1];
  Mod(dis[0][d]);
  dis[1][d] = dis[1][d * 2] + dis[1][d * 2 + 1];
  Mod(dis[1][d]);
}
signed main() {
  n = read(), q = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 1; i <= n; i++) w[i] = read();
  bt(1, n, 1);
  long long x, y, p;
  long long s, s1;
  long long ans;
  while (q--) {
    x = read(), y = read();
    if (x < 0) {
      x = -x;
      w[x] = y;
      xg(1, n, 1, x);
    } else {
      if (x == y) {
        printf("0\n");
        continue;
      }
      s = qsum(1, n, 1, x, y);
      s = (s + 1) / 2;
      if (x == 1)
        s1 = 0;
      else
        s1 = qsum(1, n, 1, 1, x - 1);
      p = kth(1, n, 1, s1 + s);
      ans = 0;
      if (p != x) {
        ans = qdis(1, n, 1, x, p - 1, 1);
        ans -= (long long)(qsum(1, n, 1, x, p - 1)) % mod *
               (a[n] - (n - p) - a[p]) % mod;
        ans %= mod;
      }
      if (p != y) {
        ans += qdis(1, n, 1, p + 1, y, 0);
        ans -= (long long)(qsum(1, n, 1, p + 1, y)) % mod * (a[p] - p) % mod;
        ans %= mod;
      }
      ans = (ans % mod + mod) % mod;
      printf("%lld\n", ans);
    }
  }
  return 0;
}