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>
void maxtt(T& t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T& t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int& x, int y, int mod = 1000000007) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
struct fw1 {
  long long b[200105];
  void update(int x, long long val) {
    for (int i = x; i <= n; i += i & -i) {
      b[i] += val;
    }
  }
  long long query(int x) {
    long long ret = 0;
    for (; x > 0; ret += b[x], x -= x & -x)
      ;
    return ret;
  }
  long long qy(int l, int r) {
    if (r < l) return 0;
    return query(r) - query(l - 1);
  }
} pw;
struct fw {
  int b[200105];
  void update(int x, int val) {
    for (int i = x; i <= n; i += i & -i) {
      addmod(b[i], val);
    }
  }
  int query(int x) {
    int ret = 0;
    for (; x > 0; addmod(ret, b[x]), x -= x & -x)
      ;
    return ret;
  }
  int qy(int l, int r) {
    if (r < l) return 0;
    return (query(r) - query(l - 1) + 1000000007) % 1000000007;
  }
} pwa;
int a[200105], w[200105], aw[200105];
void upt(int id, int nw) {
  pw.update(id, nw - w[id]);
  w[id] = nw;
  int naw = (long long)a[id] * nw % 1000000007;
  pwa.update(id, (naw + 1000000007 - aw[id]) % 1000000007);
  aw[id] = naw;
}
int cal(int l, int r) {
  if (l == r) return 0;
  long long S = pw.qy(l, r);
  int L = l, R = r;
  while (L + 1 < R) {
    int mid = (L + R) / 2;
    long long z = pw.qy(l, mid);
    if (z + z < S)
      L = mid + 1;
    else
      R = mid;
  }
  long long z = pw.qy(l, L);
  int pos;
  if (z + z < S)
    pos = R;
  else
    pos = L;
  int ans = pw.qy(l, pos - 1) % 1000000007 * a[pos] % 1000000007;
  addmod(ans, 1000000007 - pwa.qy(l, pos - 1));
  addmod(ans, pwa.qy(pos + 1, r));
  addmod(ans,
         1000000007 - pw.qy(pos + 1, r) % 1000000007 * a[pos] % 1000000007);
  return ans;
}
void fmain(int ID) {
  scanf("%d%d", &n, &m);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", a + i);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", w + i);
  for (int(i) = 1; (i) <= (int)(n); (i)++) pw.update(i, w[i]);
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    addmod(a[i], 1000000007 - i);
    aw[i] = (long long)a[i] * w[i] % 1000000007;
    pwa.update(i, aw[i]);
  }
  for (int(i) = 1; (i) <= (int)(m); (i)++) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x < 0) {
      upt(-x, y);
    } else {
      printf("%d\n", cal(x, y));
    }
  }
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}