DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
inline int scan() {
  int x = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return x;
}
long long int seg[4 * 100005] = {0};
long long int color[4 * 100005] = {0};
long long int lazy[4 * 100005] = {0};
void build(int l, int r, int pos) {
  if (l == r) {
    color[pos] = l;
  } else {
    int mid = (l + r) >> 1;
    build(l, mid, pos * 2);
    build(mid + 1, r, pos * 2 + 1);
  }
}
void checklazy(int l, int r, int pos) {
  if (!lazy[pos]) {
    return;
  }
  seg[pos] += 1LL * lazy[pos] * (r - l + 1);
  if (l != r) {
    lazy[pos * 2] += lazy[pos];
    lazy[pos * 2 + 1] += lazy[pos];
  }
  lazy[pos] = 0;
}
void godown(int l, int r, int pos, int val) {
  checklazy(l, r, pos);
  if (color[pos] > 0) {
    lazy[pos] += abs(color[pos] - val);
    checklazy(l, r, pos);
  } else {
    int mid = (l + r) >> 1;
    godown(l, mid, pos * 2, val);
    godown(mid + 1, r, pos * 2 + 1, val);
    seg[pos] = seg[pos * 2] + seg[pos * 2 + 1];
  }
}
void update(int l, int r, int pos, int ql, int qr, int val) {
  checklazy(l, r, pos);
  if (qr < l || r < ql) {
    return;
  }
  if (l >= ql && qr >= r) {
    godown(l, r, pos, val);
    color[pos] = val;
    return;
  }
  int mid = (l + r) >> 1;
  if (color[pos] > 0) {
    color[pos * 2] = color[pos];
    color[pos * 2 + 1] = color[pos];
    color[pos] = 0;
  }
  update(l, mid, pos * 2, ql, qr, val);
  update(mid + 1, r, pos * 2 + 1, ql, qr, val);
  seg[pos] = seg[pos * 2] + seg[pos * 2 + 1];
}
long long int query(int l, int r, int pos, int ql, int qr) {
  checklazy(l, r, pos);
  if (qr < l || r < ql) {
    return 0;
  }
  if (l >= ql && r <= qr) {
    return seg[pos];
  }
  int mid = (l + r) >> 1;
  return query(l, mid, pos * 2, ql, qr) +
         query(mid + 1, r, pos * 2 + 1, ql, qr);
}
int main() {
  long long int n, m;
  n = scan(), m = scan();
  build(1, n, 1);
  for (long long int i = 0; i < m; i++) {
    int type;
    long long int l, r, x;
    type = scan(), l = scan(), r = scan();
    if (type == 1) {
      x = scan();
      update(1, n, 1, l, r, x);
    } else {
      printf("%lld\n", query(1, n, 1, l, r));
    }
  }
  return 0;
}
