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;
long long sum[400009], add[400009];
int co[400009];
void pushup(int p, int l, int r) {
  sum[p] = sum[p << 1] + sum[p << 1 | 1];
  if (co[p << 1] == co[p << 1 | 1])
    co[p] = co[p << 1];
  else
    co[p] = 0;
}
void pushdown(int p, int l, int r) {
  if (add[p]) {
    add[p << 1] += add[p];
    add[p << 1 | 1] += add[p];
    int mid = (l + r) >> 1;
    sum[p << 1] += (long long)(mid - l + 1) * add[p];
    sum[p << 1 | 1] += (long long)(r - mid) * add[p];
    co[p << 1] = co[p << 1 | 1] = co[p];
    add[p] = 0;
  }
}
void build(int p, int l, int r) {
  if (l == r) {
    add[p] = sum[p] = 0;
    co[p] = l;
    return;
  }
  int mid = (l + r) >> 1;
  add[p] = 0;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
  pushup(p, l, r);
}
void update(int p, int l, int r, int ql, int qr, int d) {
  if (ql <= l && qr >= r && co[p]) {
    sum[p] += (long long)(r - l + 1) * abs(d - co[p]);
    add[p] += abs(d - co[p]);
    co[p] = d;
    return;
  }
  pushdown(p, l, r);
  int mid = (l + r) >> 1;
  if (ql <= mid) update(p << 1, l, mid, ql, qr, d);
  if (qr > mid) update(p << 1 | 1, mid + 1, r, ql, qr, d);
  pushup(p, l, r);
}
long long query(int p, int l, int r, int ql, int qr) {
  if (ql <= l && qr >= r) return sum[p];
  pushdown(p, l, r);
  int mid = (l + r) >> 1;
  long long ans = 0;
  if (ql <= mid) ans += query(p << 1, l, mid, ql, qr);
  if (qr > mid) ans += query(p << 1 | 1, mid + 1, r, ql, qr);
  pushup(p, l, r);
  return ans;
}
int main() {
  int n, i, j, k, m, ii = 0, ca;
  scanf("%d%d", &n, &m);
  build(1, 1, n);
  while (m--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int l, r, d;
      scanf("%d%d%d", &l, &r, &d);
      update(1, 1, n, l, r, d);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      cout << query(1, 1, n, l, r) << endl;
    }
  }
  return 0;
}
