192.168.0.0/22 is a /22 subnet, you can't get another /22 out of it.
Unpartitioned, this subnet would include the addresses 192.168.0.1 through 192.168.3.255 (1,024 total addresses).
Partitioning the subnet into smaller subnets will solve your problem. You cannot divide networks into precise arbitrary divisions like 190 addresses or 55 addresses; they must be divided into even blocks like 2x 512 addresses, 4x 256 addresses, 8x 128 addresses, etc., or any mix of sizes that adds up to 1024 (in this case of subdividing a /22).
So I think you should create subnets that are large enough to provide the required number of addresses, for example,
190 Hosts
192.168.0.0/24
Address Space: 192.168.0.0 - 192.168.0.255
Gateway: 192.168.0.1
Available WAN IPs: 192.168.0.2 - 192.168.0.254 (252 Hosts)
55 Hosts
192.168.1.0/26
Address Space: 192.168.1.0 - 192.168.1.64
Gateway: 192.168.1.1
Available WAN IPs: 192.168.1.2 - 192.168.1.62 (60 Hosts)
26 Hosts
192.168.1.65/27
Address Space: 192.168.1.65 - 192.168.1.96
Gateway: 192.168.1.66
Available WAN IPs: 192.168.1.67 - 192.168.1.94 (28 Hosts)
This leaves the remainder free, which could be divided into the following additional subnets to fully utilize the original /22 net.
1x /27 (32 addresses)
1x /25 (128 addresses)
1x /23 (512 addresses)
Note that "available WAN IPs" and # of hosts may vary slightly, depending on how many addresses are used for management functions on this network (minimum is 2, gateway first address and loopback last address).
Source: Your best friend the CIDR table http://www.rjsmith.com/CIDR-Table.html