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 <class T>
int chkmax(T& a, T b) {
  if (b > a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
int chkmin(T& a, T b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
template <class iterator>
void output(iterator begin, iterator end, ostream& out = cerr) {
  while (begin != end) {
    out << (*begin) << " ";
    begin++;
  }
  out << '\n';
}
template <class T>
void output(T x, ostream& out = cerr) {
  output(x.begin(), x.end(), out);
}
void fast_io() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long MOD = 1e9 + 7;
struct data {
  long long val, cnt, sum;
  data() : val(0), cnt(0), sum(0) {}
  data(long long _x) : val(0), cnt(1), sum(_x) {}
  data(long long _val, long long _cnt, long long _sum)
      : val(_val), cnt(_cnt), sum(_sum) {}
};
data merge(data a, data b) {
  return data((a.val + b.val + b.sum * a.cnt) % MOD, a.cnt + b.cnt,
              (a.sum + b.sum) % MOD);
}
struct st {
  vector<data> t;
  int n;
  void init(int _n) {
    n = _n;
    t.resize(4 * n);
  }
  void update(int pos, long long val, int v, int tl, int tr) {
    if (tl == tr) {
      t[v] = data(val);
    } else {
      int tm = (tl + tr) >> 1;
      if (pos <= tm)
        update(pos, val, v * 2, tl, tm);
      else
        update(pos, val, v * 2 + 1, tm + 1, tr);
      t[v] = merge(t[v * 2], t[v * 2 + 1]);
    }
  }
  void output(int v, int tl, int tr) {
    cerr << tl << ".." << tr << ": " << t[v].val << " " << t[v].cnt << " "
         << t[v].sum << '\n';
    if (tl < tr) {
      int tm = (tl + tr) >> 1;
      output(v * 2, tl, tm);
      output(v * 2 + 1, tm + 1, tr);
    }
  }
  data get(int l, int r, int v, int tl, int tr) {
    if (l > r) return data();
    if (l == tl && r == tr) return t[v];
    int tm = (tl + tr) >> 1;
    data res_l = get(l, min(r, tm), v * 2, tl, tm);
    data res_r = get(max(l, tm + 1), r, v * 2 + 1, tm + 1, tr);
    return merge(res_l, res_r);
  }
  void update(int pos, long long val) { update(pos, val, 1, 0, n - 1); }
  long long get(int l, int r) {
    data x = get(l, r, 1, 0, n - 1);
    return x.val;
  }
};
struct fenwick_mod {
  vector<long long> t;
  int n;
  void init(int _n) {
    n = _n;
    t.resize(n);
  }
  void update(int pos, long long val) {
    while (pos < n) {
      t[pos] = (t[pos] + val) % MOD;
      pos = (pos | (pos + 1));
    }
  }
  long long get(int r) {
    long long res = 0;
    while (r >= 0) {
      res += t[r];
      r = (r & (r + 1)) - 1;
    }
    return res % MOD;
  }
  long long get(int l, int r) { return (get(r) - get(l - 1) + MOD) % MOD; }
};
struct fenwick {
  vector<long long> t;
  int n;
  void init(int _n) {
    n = _n;
    t.resize(n);
  }
  void update(int pos, long long val) {
    while (pos < n) {
      t[pos] = (t[pos] + val);
      pos = (pos | (pos + 1));
    }
  }
  long long get(int r) {
    long long res = 0;
    while (r >= 0) {
      res += t[r];
      r = (r & (r + 1)) - 1;
    }
    return res;
  }
  long long get(int l, int r) { return get(r) - get(l - 1); }
};
const int mx = 2e5 + 228;
int n, q;
long long a[mx], w[mx];
fenwick_mod f_aw;
fenwick f_w;
st t, t_rev;
void update_index(int pos, long long new_w, long long old_w) {
  long long diff = (new_w - old_w + MOD) % MOD;
  f_w.update(pos, new_w - old_w);
  f_aw.update(pos, (a[pos] * diff) % MOD);
  t.update(pos, new_w);
  t_rev.update(n - pos - 1, new_w);
}
void init() {
  f_w.init(n);
  f_aw.init(n);
  t.init(n);
  t_rev.init(n);
  for (int i = 0; i < n; ++i) {
    update_index(i, w[i], 0);
  }
}
void update_query(int pos, long long new_w) {
  update_index(pos, new_w, w[pos]);
  w[pos] = new_w;
}
int get_max_left_pos(int l0, int r0, long long L) {
  int l = l0 - 1, r = r0;
  while (l != r) {
    int mid = (l + r) >> 1;
    if (a[mid + 1] < L + (long long)(mid + 1 - l0)) {
      l = mid + 1;
    } else {
      r = mid;
    }
  }
  return l;
}
int get_min_right_pos(int l0, int r0, long long L) {
  int l = l0, r = r0 + 1;
  while (l != r) {
    int mid = (l + r) >> 1;
    if (a[mid] > L + (long long)(mid - l0))
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
long long to_mod(long long x) { return (x % MOD + MOD) % MOD; }
long long get_l(int l0, int r0, long long R) {
  long long res = to_mod(R * (f_w.get(l0, r0) % MOD) - f_aw.get(l0, r0) -
                         t_rev.get(n - 1 - r0, n - 1 - l0));
  return res;
}
long long get_r(int l0, int r0, long long L) {
  long long res =
      to_mod(f_aw.get(l0, r0) - (f_w.get(l0, r0) % MOD) * L - t.get(l0, r0));
  return res;
}
long long get(int l0, int r0, long long L) {
  int mid_pos = get_max_left_pos(l0, r0, L);
  long long res = to_mod(get_l(l0, mid_pos, L + mid_pos - l0) +
                         get_r(mid_pos + 1, r0, L + mid_pos + 1 - l0));
  return res;
}
long long get_query(int l0, int r0) {
  long long len = (long long)(r0 - l0 + 1), L = a[l0], R = a[r0] - len + 1;
  while (R > L) {
    long long MID = (L + R) >> 1;
    int max_left_pos = get_max_left_pos(l0, r0, MID),
        min_right_pos = get_min_right_pos(l0, r0, MID);
    long long wl = f_w.get(l0, max_left_pos), wr = f_w.get(min_right_pos, r0),
              wm = f_w.get(max_left_pos + 1, min_right_pos - 1),
              diff_to_next = wl - wr + wm;
    if (diff_to_next >= 0)
      R = MID;
    else
      L = MID + 1;
  }
  long long ans = get(l0, r0, L);
  return ans % MOD;
}
signed main() {
  fast_io();
  cin >> n >> q;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> w[i];
  }
  init();
  for (int i = 0; i < q; ++i) {
    int id;
    long long val;
    cin >> id >> val;
    if (id < 0) {
      update_query(-id - 1, val);
    } else {
      cout << get_query(id - 1, (int)val - 1) << '\n';
    }
  }
}