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]. 
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long _INF = 0xc0c0c0c0c0c0c0c0;
const long long mod = (int)1e9 + 7;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long ksm(long long a, long long b, long long mod) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long inv2(long long a, long long mod) { return ksm(a, mod - 2, mod); }
void exgcd(long long a, long long b, long long &x, long long &y, long long &d) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    exgcd(b, a % b, y, x, d);
    y -= x * (a / b);
  }
}
const int MAX_N = 200025;
long long s[MAX_N << 2], sum[MAX_N << 2], arr[MAX_N], b[MAX_N], w[MAX_N];
int n;
void up(int rt) {
  sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
  s[rt] = s[rt << 1] + s[rt << 1 | 1];
}
void build(int rt, int l, int r) {
  sum[rt] = s[rt] = 0;
  if (l == r) {
    sum[rt] = w[l];
    s[rt] = b[l] * w[l] % 1000000007;
    return;
  }
  build(rt << 1, l, ((l + r) >> 1));
  build(rt << 1 | 1, ((l + r) >> 1) + 1, r);
  up(rt);
}
void update(int rt, int l, int r, int x, int v) {
  if (l == r) {
    sum[rt] = v;
    s[rt] = 1ll * b[x] * v % 1000000007;
    return;
  }
  if (x <= ((l + r) >> 1))
    update(rt << 1, l, ((l + r) >> 1), x, v);
  else
    update(rt << 1 | 1, ((l + r) >> 1) + 1, r, x, v);
  up(rt);
}
long long query(int rt, int l, int r, int x, int y) {
  if (x > r || y < l) return 0;
  if (x <= l && r <= y) {
    return sum[rt];
  }
  if (x > ((l + r) >> 1))
    return query(rt << 1 | 1, ((l + r) >> 1) + 1, r, x, y);
  else if (y <= ((l + r) >> 1))
    return query(rt << 1, l, ((l + r) >> 1), x, y);
  else
    return query(rt << 1, l, ((l + r) >> 1), x, y) +
           query(rt << 1 | 1, ((l + r) >> 1) + 1, r, x, y);
}
long long ask(int rt, int l, int r, int x, int y) {
  if (x > r || y < l) return 0;
  if (x <= l && r <= y) {
    return s[rt];
  }
  if (x > ((l + r) >> 1))
    return ask(rt << 1 | 1, ((l + r) >> 1) + 1, r, x, y);
  else if (y <= ((l + r) >> 1))
    return ask(rt << 1, l, ((l + r) >> 1), x, y);
  else
    return (ask(rt << 1, l, ((l + r) >> 1), x, y) % 1000000007 +
            ask(rt << 1 | 1, ((l + r) >> 1) + 1, r, x, y) % 1000000007) %
           1000000007;
}
long long solve(int x, int y) {
  int l = x, r = y, pos = l;
  while (l <= r) {
    long long tmp = query(1, 1, n, x, ((l + r) >> 1)),
              tmp_ = query(1, 1, n, ((l + r) >> 1) + 1, y);
    if (tmp >= tmp_)
      pos = ((l + r) >> 1), r = ((l + r) >> 1) - 1;
    else
      l = ((l + r) >> 1) + 1;
  }
  long long tmp = (query(1, 1, n, x, pos - 1) % 1000000007 *
                       ((arr[pos] - pos + 1000000007) % 1000000007) -
                   ask(1, 1, n, x, pos - 1) + 1000000007) %
                  1000000007;
  long long tmp_ = (query(1, 1, n, pos + 1, y) % 1000000007 *
                        ((pos - arr[pos] + 1000000007) % 1000000007) +
                    ask(1, 1, n, pos + 1, y) % 1000000007) %
                   1000000007;
  return ((tmp + tmp_) % 1000000007);
}
int main() {
  int Q, x;
  long long y;
  scanf("%d%d", &n, &Q);
  for (int i = 1; i <= n; ++i) scanf("%lld", &arr[i]), b[i] = arr[i] - i;
  for (int i = 1; i <= n; ++i) scanf("%lld", &w[i]);
  build(1, 1, n);
  while (Q--) {
    scanf("%d%lld", &x, &y);
    if (x < 0) {
      update(1, 1, n, -x, y);
    } else {
      printf("%lld\n", solve(x, y));
    }
  }
  return 0;
}
