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;
const int MAXN = 111111;
long long n, m, a[MAXN], cnt, last[MAXN], tree[MAXN << 2], lazy[MAXN << 2],
    color[MAXN << 2];
inline void build(const int &x, const int &l, const int &r) {
  tree[x] = lazy[x] = 0;
  color[x] = -1;
  if (l == r) {
    color[x] = l;
    return;
  }
  build((x << 1), l, ((l + r) >> 1));
  build(((x << 1) | 1), ((l + r) >> 1) + 1, r);
}
inline void pushdown(const int &x, const int &l, const int &r) {
  if (lazy[x]) {
    lazy[(x << 1)] += lazy[x];
    lazy[((x << 1) | 1)] += lazy[x];
    tree[(x << 1)] += lazy[x] * (((l + r) >> 1) - l + 1);
    tree[((x << 1) | 1)] += lazy[x] * (r - ((l + r) >> 1));
    color[(x << 1)] = color[((x << 1) | 1)] = color[x];
    lazy[x] = 0;
  }
}
inline void pushup(const int &x) {
  if (color[(x << 1)] == color[((x << 1) | 1)] && color[(x << 1)] != -1)
    color[x] = color[(x << 1)];
  else
    color[x] = -1;
  tree[x] = tree[(x << 1)] + tree[((x << 1) | 1)];
}
inline void modify(const int &x, const int &l, const int &r, const int &ql,
                   const int &qr, const long long &v) {
  if (l != r) pushdown(x, l, r);
  if (l == ql && r == qr && color[x] != -1) {
    long long tmp = abs(v - color[x]);
    tree[x] += tmp * (r - l + 1);
    color[x] = v;
    lazy[x] += tmp;
    return;
  }
  if (qr <= ((l + r) >> 1))
    modify((x << 1), l, ((l + r) >> 1), ql, qr, v);
  else if (ql > ((l + r) >> 1))
    modify(((x << 1) | 1), ((l + r) >> 1) + 1, r, ql, qr, v);
  else {
    modify((x << 1), l, ((l + r) >> 1), ql, ((l + r) >> 1), v);
    modify(((x << 1) | 1), ((l + r) >> 1) + 1, r, ((l + r) >> 1) + 1, qr, v);
  }
  pushup(x);
}
inline long long query(const int &x, const int &l, const int &r, const int &ql,
                       const int &qr) {
  if (l != r) pushdown(x, l, r);
  if (l == ql && r == qr) return tree[x];
  if (qr <= ((l + r) >> 1))
    return query((x << 1), l, ((l + r) >> 1), ql, qr);
  else if (ql > ((l + r) >> 1))
    return query(((x << 1) | 1), ((l + r) >> 1) + 1, r, ql, qr);
  return query((x << 1), l, ((l + r) >> 1), ql, ((l + r) >> 1)) +
         query(((x << 1) | 1), ((l + r) >> 1) + 1, r, ((l + r) >> 1) + 1, qr);
}
int main() {
  scanf("%I64d%I64d\n", &n, &m);
  build(1, 1, n);
  while (m--) {
    int cmd, l, r;
    long long x;
    scanf("%d", &cmd);
    if (cmd == 1) {
      scanf("%d%d%I64d", &l, &r, &x);
      modify(1, 1, n, l, r, x);
    } else {
      scanf("%d%d", &l, &r);
      cout << query(1, 1, n, l, r) << endl;
    }
  }
}
