The year 2015 is almost over.

Limak is a little polar bear. He has recently learnt about the binary system. He noticed that the passing year has exactly one zero in its representation in the binary system — 2015<sub class="lower-index">10</sub> = 11111011111<sub class="lower-index">2</sub>. Note that he doesn't care about the number of zeros in the decimal representation.

Limak chose some interval of years. He is going to count all years from this interval that have exactly one zero in the binary representation. Can you do it faster?

Assume that all positive integers are always written without leading zeros.
The only line of the input contains two integers *a* and *b* (1 ≤ *a* ≤ *b* ≤ 10<sup class="upper-index">18</sup>) — the first year and the last year in Limak's interval respectively.
In the first sample Limak's interval contains numbers 5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>, 6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>, 7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>, 8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>, 9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub> and 10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>. Two of them (101<sub class="lower-index">2</sub> and 110<sub class="lower-index">2</sub>) have the described property.
5 10
2
2015 2015
1
100 105
0
72057594000000000 72057595000000000
26
1 100
16
1000000000000000000 1000000000000000000
0
1 1000000000000000000
1712
1 1
0
1 2
1
1 3
1
1 4
1
1 5
2
1 6
3
1 7
3
2 2
1
2 3
1
2 4
1
2 5
2
2 6
3
2 7
3
3 3
0
3 4
0
3 5
1
3 6
2
3 7
2
4 4
0
4 5
1
4 6
2
4 7
2
5 5
1
5 6
2
5 7
2
6 6
1
6 7
1
7 7
0
1 8
3
6 8
1
7 8
0
8 8
0
1 1022
45
1 1023
45
1 1024
45
1 1025
45
1 1026
45
509 1022
11
510 1022
10
511 1022
9
512 1022
9
513 1022
9
509 1023
11
510 1023
10
511 1023
9
512 1023
9
513 1023
9
509 1024
11
510 1024
10
511 1024
9
512 1024
9
513 1024
9
509 1025
11
510 1025
10
511 1025
9
512 1025
9
513 1025
9
1 1000000000
408
10000000000 70000000000000000
961
1 935829385028502935
1712
500000000000000000 1000000000000000000
58
500000000000000000 576460752303423488
57
576460752303423488 1000000000000000000
1
999999999999999999 1000000000000000000
0
1124800395214847 36011204832919551
257
1124800395214847 36011204832919550
256
1124800395214847 36011204832919552
257
1124800395214846 36011204832919551
257
1124800395214848 36011204832919551
256
1 287104476244869119
1603
1 287104476244869118
1602
1 287104476244869120
1603
492581209243647 1000000000000000000
583
492581209243646 1000000000000000000
583
492581209243648 1000000000000000000
582
1099444518911 1099444518911
1
1099444518910 1099444518911
1
1099444518911 1099444518912
1
1099444518910 1099444518912
1
864691128455135231 864691128455135231
1
864691128455135231 864691128455135232
1
864691128455135230 864691128455135232
1
864691128455135230 864691128455135231
1
864691128455135231 1000000000000000000
1
864691128455135232 1000000000000000000
0
864691128455135230 1000000000000000000
1
576460752303423487 576460752303423487
0
1 576460752303423487
1711
1 576460752303423486
1711
2 1000000000000000000
1712
3 1000000000000000000
1711
4 1000000000000000000
1711
5 1000000000000000000
1711
6 1000000000000000000
1710
5 6
2
1 2
1
x, y = map(int, input().split())
ans = 0
for left in range(1, 100):
    s = '1' * left + '0'
    for right in range(100):
        a = int(s, 2)
        if a in range(x,y+1):
          ans += 1
        s += '1'
print(ans)